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\title{Circle geometry }

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\maketitle
You have already seen how powerful Euclidean geometry is when working with triangles. For example, Pythagoras' theorem and all of trigonometry arise from Euclidean geometry.

When applied to circles, geometry also produces beautiful and surprising results. In this chapter, you will see how useful congruence and similarity are in the context of circle geometry.

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-01}
Photocopying is restricted under law and this material must not be transferred to another party

\section*{13 Angles at the centre and the circumference}
A circle is the set of all points that lie a fixed distance (called the radius) from a fixed point (called the centre).

While we use the word 'radius' to mean this fixed distance, we also use 'radius' to mean any interval joining a point on the circle to the centre. The radii (plural of radius) of a circle radiate out from the centre, like the spokes of a bicycle wheel. (The word radius is a Latin word meaning 'spoke' or 'ray'.)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-02(1)}
\end{center}

A chord of a circle is the interval joining any two points on the circle. The word chord is a Greek word meaning 'cord' or 'string'. A tightly stretched string that is plucked gives out a musical note, which is the origin of the word chord in music.

A chord that passes through the centre of the circle is called a diameter.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-02(2)}
\end{center}

\section*{Angles in a semicircle}
We will start this chapter with an important result about circles. The discovery of this result was attributed to Thales ( \(600 \mathrm{BC}\) ) by later Greek mathematicians, who claimed that it was the first theorem ever consciously stated and proved in mathematics.

In each diagram below, the angle \(\angle P\) is called an angle in a semicircle. It is formed by taking a diameter \(A O B\), choosing any other point \(P\) on the circle, and joining the chords \(P A\) and \(P B\).
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-02(3)}

These diagrams lead us to ask the question, 'What happens to \(\angle P\) as \(P\) takes different positions around the semicircle?' Thales discovered a marvellous fact: \(\angle P\) is always a right angle.

\section*{Theorem:}
Proof:
An angle in a semicircle is a right angle.

Draw the radius \(O P\).

Because the radii are equal, we have two isosceles triangles \(\triangle A O P\) and \(\triangle B O P\).

Let \(\angle B A P=\alpha\) and \(\angle A B P=\beta\).

Then \(\angle O P A=\alpha\) (base angles of isosceles \(\triangle O P A)\) and \(\angle O P B=\beta\) (base angles of isosceles \(\triangle O P B\) ).

Adding up the interior angles of the triangle \(\triangle A B P\),

\[
\begin{aligned}
\alpha+\alpha+\beta+\beta & =180^{\circ} \\
\alpha+\beta & =90^{\circ}
\end{aligned}
\]

So

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-02}
\end{center}

\section*{Angles in a semicircle (Thales' theorem)}
An angle in a semicircle is a right angle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-03(1)}
\end{center}

Two slightly different proofs of this result are given in Exercise 13A.

\section*{Example 1}
In the diagram shown, \(O\) is the centre of the circle.

a Find \(\alpha\).

b Prove that \(A P B Q\) is a rectangle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-03}
\end{center}

\section*{Solution}
a First, \(\angle P=90^{\circ}\) (angle in a semicircle) so \(\alpha=65^{\circ}\) (angle sum of \(\triangle A P B\) )

b Also, \(\angle Q=90^{\circ}\) (angle in a semicircle) so \(\angle P A Q=90^{\circ}\) and \(\angle P B Q=90^{\circ}\) (co-interior angles, \(A Q \| B P\) ) so \(A P B Q\) is a rectangle, being a quadrilateral with interior angles that are all \(90^{\circ}\).

\section*{Arcs and segments}
Our next result needs some additional words.

A chord \(A B\) divides the circle into two regions called segments.

If \(A B\) is not a diameter, the regions are unequal. The larger region is called the major segment and the smaller region is called the minor segment.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-03(2)}
\end{center}

Similarly, the points \(A\) and \(B\) divide the circumference into two pieces called arcs.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-03(3)}
\end{center}

If \(A B\) is not a diameter, the arcs are unequal. The larger piece is called the major arc and the smaller piece is called the minor arc. Notice that the phrase 'the arc \(A B\) ' could refer to either arc, and we often need to clarify which arc we mean.

\section*{Angles at the centre and the circumference}
Thales' theorem about an angle in a semicircle is a special case of a more general result.

Consider a fixed arc \(A F B\). Consider a point \(P\) on the other arc. Join the chords \(A P\) and \(B P\) to form the angle \(\angle A P B\).

We call this angle \(\angle A P B\) an angle at the circumference subtended by the arc \(A F B\). The word subtends comes from Latin and literally means 'stretches under' or 'holds under'. We also say \(\angle A P B\) stands on the arc \(A F B\).

As with angles in a semicircle, we ask, 'What happens to \(\angle A P B\) as \(P\) takes different positions around the arc?'

We will begin our study by focusing on \(\angle A P B\) in relation to \(\angle A O B\), the angle subtended at the centre of the circle by the arc \(A F B\). There are four cases to consider.

\begin{center}
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\end{center}

Case 1

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\end{center}

Case 2

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\end{center}

Case 3

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\end{center}

Case 4

In the first three cases, \(A F B\) is a minor arc and \(\angle A P B\) is acute. However, in the fourth case, \(A F B\) is a major arc and \(\angle A P B\) is obtuse.

\section*{Case 1}
Suppose that \(A F B\) is a minor arc, and \(A, B\) and \(P\) are located as in the diagram. We draw all three radii, \(A O, B O\) and \(P O\), and produce \(P O\) to \(X\).

Since \(A O\) and \(P O\) are radii, \(\triangle A O P\) is isosceles. Let the equal angles be \(\alpha\).

Similarly, \(\triangle B O P\) is isosceles. Let the equal angles be \(\beta\).

Next, using the fact that an exterior angle of a triangle is equal to the sum of the

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-04(2)}
two opposite interior angles, \(\angle A O X=2 \alpha\) and \(\angle B O X=2 \beta\).

Hence, \(\angle A O B=2 \alpha+2 \beta\) and \(\angle A P B=\alpha+\beta\). The following result has now been proved for case 1 .

Theorem: The angle at the centre subtended by an arc of a circle is twice an angle at the circumference subtended by the same arc.

The proof for Case 4 is the same as the above proof for Case 1. It relates the obtuse angle \(\angle A P B\) to the reflex angle \(\angle A O B\). The other two cases will be dealt with in question 7 of Exercise 13A. This will complete the proof of the theorem. Some examples of the relationship between angles at the circumference and at the centre, when subtended by a common arc, are illustrated in the following diagrams.
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-04(5)}

\section*{Semicircles}
As we can see from the fourth example from the previous page, when the arc is a semicircle, the angle at the centre is \(180^{\circ}\) and the angle at the circumference is \(90^{\circ}\). You will recognise that this is precisely the situation covered by Thales' theorem, which is thus a special case of our new theorem. Thus, the two theorems are an excellent example of a theorem and its generalisation. This situation occurs routinely throughout mathematics. For example, the cosine rule can be thought of as a generalisation of Pythagoras' theorem.

\section*{Angles at the centre and the circumference}
The angle at the centre subtended by an arc of a circle is twice an angle at the circumference subtended by the same arc.

\section*{Example 2}
Find \(\alpha\) and \(\beta\) in the diagram shown, where \(O\) is the centre of the circle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-05}
\end{center}

\section*{Solution}
\(\alpha=60^{\circ}\) (angle at the centre is half the angle at the circumference on the same arc \(A Q B\) )

Next, reflex \(\angle A O B=240^{\circ}(\) angles in a revolution at \(O\) )

so \(\beta=120^{\circ}\) (angle at the centre is half the angle at the circumference on the same arc \(A P B\) )

\section*{Exercise 13A}
Note: Points labelled \(O\) in this exercise are always centres of circles.

1 a i Use compasses to draw a large circle with centre \(O\), and draw a diameter \(A O B\).

ii Draw an angle \(\angle A P B\) in one of the semicircles. What is its size?

b i Draw another large circle, and draw a chord \(A B\) that is not a diameter.

ii Draw the angle at the centre and an angle at the circumference subtended by the minor arc \(A B\). How are these two angles related?

iii Mark the angle at the centre on the major arc \(A B\) and draw an angle at the circumference subtended by this major arc. How are these two angles related?

2 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons.

a

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c

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e

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b

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\end{center}

d

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\end{center}

f

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\end{center}

3 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons.

a

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\end{center}

d

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g

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\end{center}

b

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\end{center}

e

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\end{center}

h

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\end{center}

c

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\end{center}

f

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\end{center}

i

\begin{center}
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\end{center}

4 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-07(4)}
\end{center}

d

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\end{center}

g

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\end{center}

b

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\end{center}

e

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\end{center}

h

\begin{center}
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\end{center}

c

\begin{center}
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\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-07(3)}
\end{center}

i
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-07(13)}

c

e
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-07(5)}

b

\(\mathbf{f}\)

\begin{center}
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\end{center}

d

\begin{center}
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\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-07(2)}
\end{center}

6 Thales' theorem states that: An angle in a semicircle is a right angle. This question develops two other proofs of Thales' theorem. We must prove, in each part, that \(\angle A P B=90^{\circ}\).

a (Euclid's proof) Join \(P O\), and produce \(A P\) to \(X\).

Let \(\angle A=\alpha\) and \(\angle B=\beta\).

i Prove that \(\angle A P B=\alpha+\beta\), and that \(\angle X P B=\alpha+\beta\).

ii Hence, prove that \(\alpha+\beta=90^{\circ}\).

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-07(11)}
b Join \(P O\) and produce it to \(M\).

i Prove that \(\angle A O M=2 \alpha\) and \(\angle B O M=2 \beta\).

ii Hence, prove that \(2 \alpha+2 \beta=180^{\circ}\).

iii Deduce that \(\alpha+\beta=90^{\circ}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-08(3)}
\end{center}

7 Prove that: An angle at the centre subtended by an arc is twice an angle at the circumference subtended by the same arc.

We proved Case 1 of this and noted Case 4 follows in the same manner. We pointed out that there are two other cases to consider.

a Using the diagram to the right:

i prove that \(\angle A P B=\beta\)

ii prove that \(\angle A O B=2 \beta\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-08}
\end{center}

b Using the diagram to the right:

i prove that \(\angle A P B=\beta-\alpha\)

ii prove that \(\angle A O B=2(\beta-\alpha)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-08(2)}
\end{center}

8 The converse of Thales' theorem is established by proving the following result:

The midpoint of the hypotenuse of a right-angled triangle is equidistant from the three vertices of the triangle.

Let \(\triangle A B P\) be right-angled at \(P\), and let \(O\) be the midpoint of the hypotenuse \(A B\). Draw \(P O\) and produce it to \(Q\) so that \(P O=O Q\). Draw \(A Q\) and \(B Q\).

a Explain why \(A P B Q\) is a parallelogram.

b Hence, explain why \(A P B Q\) is a rectangle.

c Hence, explain why \(A O=B O=P O\) and why the circle with

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-08(1)}
diameter \(A B\) passes through \(P\).

9 (An application of the angle at the centre and circumference theorem) A horse is travelling around a circular track at a constant speed. A punter standing at the very edge of the track is following him with binoculars. Explain why the punter's binoculars are rotating at a constant rate.

\section*{Angles at the circumference and cyclic quadrilaterals}
Let us look at three angles at the circumference, all subtended by the same arc \(A F B\).
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-09(2)}

We know already that all three angles are half the angle \(\angle A O B\) at the centre subtended by this same arc.

It follows immediately that all three angles are equal. This result is important enough to state as a separate theorem, in the box below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-09}
\end{center}

\section*{Angles at the circumference}
Angles at the circumference of a circle subtended by the same arc are equal.

This is often stated as 'Angles in the same segment are equal'.

\section*{Example 3}
Find \(\alpha, \beta\) and \(\gamma\) in the diagram to the right.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-09(1)}
\end{center}

\section*{Solution}
First, \(\alpha=60^{\circ}\) (angles on the same arc \(A B\) )

Second, \(\beta=20^{\circ}\) (angles on the same arc \(P Q\) )

Third, \(\gamma=80^{\circ}\) (exterior angle of \(\triangle A P M\) )

\section*{Cyclic quadrilaterals}
A cyclic quadrilateral is a quadrilateral whose vertices all lie on a circle.

We also say that the points \(A, B, P\) and \(Q\) are concyclic.

The opposite angles \(\angle P\) and \(\angle Q\) of the cyclic quadrilateral \(A P B Q\) are closely related.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-10}
\end{center}

The key to finding the relationship is to draw the radii \(A O\) and \(B O\).

First, \(\angle P\) is half the angle \(\angle A O B\) at the centre on the same arc \(A Q B\); we have marked these angles \(\alpha\) and \(2 \alpha\).

Second, \(\angle Q\) is half the reflex angle \(\angle A O B\) at the centre on the same arc \(A P B\); we have marked these angles \(\beta\) and \(2 \beta\).

SO

\[
\begin{aligned}
2 \alpha+2 \beta & =360^{\circ} \quad(\text { angles in a revolution at } O) \\
\alpha+\beta & =180^{\circ}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-10(1)}
\end{center}

Hence, the opposite angles \(\angle P\) and \(\angle Q\) are supplementary.

The diagram could also have been drawn as shown, but the proof is unchanged.

We usually state this as a result about cyclic quadrilaterals.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-10(3)}
\end{center}

\section*{Cyclic quadrilaterals}
The opposite angles of a cyclic quadrilateral are supplementary.

An interesting alternative proof is given as question \(\mathbf{8}\) in Exercise 13B. Every cyclic quadrilateral is convex because none of its angles are reflex, but not every convex quadrilateral is cyclic.

\section*{Example 4}
Find \(\alpha, \beta\) and \(\gamma\) in the diagram shown.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-10(2)}
\end{center}

\section*{Solution}
\[
\begin{aligned}
\alpha+70^{\circ} & =180^{\circ}(\text { opposite angles of cyclic quadrilateral } A B C D) \\
\alpha & =110^{\circ} \\
\beta+100^{\circ} & =180^{\circ}(\text { opposite angles of cyclic quadrilateral } A B C D) \\
\beta & =80^{\circ} \\
\gamma+\beta & =180^{\circ}(\text { straight angle at } B) \\
\gamma & =100^{\circ}
\end{aligned}
\]

\section*{The converse of Thales' theorem}
We began this chapter by proving Thales' theorem:

An angle in a semicircle is a right angle.

Thales' theorem has an important converse, which was proved in question 8 of exercise 13A. It elegantly uses the diagonal properties of rectangles.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-11(1)}
\end{center}

\section*{Converse of Thales' theorem}
If an interval \(A B\) subtends a right angle at a point \(P\), then \(P\) lies on the circle with diameter \(A B\).

Here is a diagram that illustrates the converse of Thales' theorem very nicely. Suppose that \(A B\) is a line interval. A person walks from \(A\) to \(B\) in a curved path \(A P Q R S B\) so that \(A B\) always subtends a right angle at his position.

The converse of Thales' theorem tells us that his path is a semicircle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-11}
\end{center}

\section*{Exercise 13B}
Note: Points labelled \(O\) in this exercise are always centres of circles.

1 a Draw a large circle, and draw a chord \(A B\) that is not a diameter.

b Draw two angles at the circumference standing on the minor arc \(A B\). How are these two angles related?

c Draw two angles at the circumference standing on the major arc \(A B\). How are these two angles related?

2 Find the values of \(\alpha, \beta\) and \(\theta\), giving reasons.

a

\begin{center}
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\end{center}

d

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\end{center}

b

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\end{center}

\(\mathbf{e}\)

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\end{center}

c

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\end{center}

f

\begin{center}
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\end{center}

3 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-12}
\end{center}

d

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\end{center}

b

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\end{center}

e

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\end{center}

c

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f

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\end{center}

c

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d
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e

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\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-12(15)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-12(8)}
\end{center}

ICE-EM MATHEMATICS YEAR 10 ICE-EM Mathematics 10 3ed ISBN 978-1-108-40434-1 C The University of Melbourne / AMSI 2017
Photocopying is restricted under law and this material must not be transferred to another party

5 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-13(4)}
\end{center}

i Prove that

\[
\angle P=\angle Q=\angle S=\angle T \text {. }
\]

ii Prove that \(P T=S Q\). b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-13}
\end{center}

i Find \(\alpha, \beta\) and \(\gamma\).

ii Prove that \(P Q \perp G R\).

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-13(1)}
\end{center}

i Find \(\alpha, \beta\) and \(\gamma\).

ii Prove that \(A B \| P Q\).

iii Prove that \(A P=B Q\).

7 The centres of the circles are \(O\) and \(P\).

a i Find \(\angle A B S\) and \(\angle A B T\).

ii Hence, prove that \(S, B\) and \(T\) are collinear.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-13(5)}
\end{center}

b i Find \(\angle A B C\) and \(\angle A B D\).

ii Hence, prove that \(C, B\) and \(D\) are collinear.

iii Why is \(A C\) a diameter?

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-13(3)}
\end{center}

8 Here is an alternative proof that: The opposite angles of a cyclic quadrilateral are supplementary.

In the cyclic quadrilateral \(A B C D\), draw the diagonals \(A C\) and \(B D\).

Let \(\alpha=\angle D A C\) and \(\beta=\angle B A C\).

a Prove that \(\angle D B C=\alpha\) and \(\angle B D C=\beta\).

b Hence, prove that \(\angle D C B=180^{\circ}-(\alpha+\beta)\).

c Deduce that \(\angle D A B+\angle D C B=180^{\circ}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-13(2)}
\end{center}

9 a Prove that: A cyclic parallelogram is a rectangle.

In the cyclic parallelogram \(A B C D\), let \(\angle A=\theta\).

i Give reasons why \(\angle C=180^{\circ}-\theta\) and why \(\angle C=\theta\).

ii Hence, prove that \(A B C D\) is a rectangle.

b Use part a to prove that: A cyclic rhombus is a square.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-14}
\end{center}

10 Prove that: In a cyclic trapezium that is not a parallelogram, the non-parallel sides have equal length.

Let \(A B C D\) be a cyclic trapezium with \(A B \| D C\) and \(A D \not \| B C\). Suppose \(D A\) meets \(C B\) at \(M\) and let \(\angle D A B=\theta\).

a Prove that \(\triangle A B M\) and \(\triangle D C M\) are isosceles.

b Hence, prove that \(A D=B C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-14(3)}
\end{center}

\section*{13 Chords and angles at the centre}
Take a minor arc \(A F B\) of a circle and join the radii \(A O\) and \(B O\). The angle \(\angle A O B\) at the centre is called the angle subtended at the centre by the arc \(A F B\). It is also called the angle at the centre subtended by the chord \(A B\).

\[
O A=O B \text { (radii of a circle) }
\]

Therefore, \(\triangle A O B\) is isosceles. This is the key idea in the results of this section.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-14(1)}
\end{center}

\section*{Equal chords and equal angles at the centre}
Suppose now that two chords each subtend an angle at the centre of the circle. Two results about this situation can be proved.

Theorem: Chords of equal length subtend equal angles at the centre of the circle.

Proof: \(\quad\) In the diagram, \(A B\) and \(P Q\) are chords of equal length.

\[
O A=O B=O P=O Q \text { (radii of a circle) }
\]

From the diagram,

\[
\triangle A O B \equiv \triangle P O Q(\mathrm{SSS})
\]

Hence, \(\angle A O B=\angle P O Q\) (matching angles of congruent triangles)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-14(2)}
\end{center}

Theorem: Conversely, chords subtending equal angles at the centre have equal length.

Proof: \(\quad\) In the diagram, \(A B\) and \(P Q\) subtend equal angles at \(O\),

so \(\triangle A O B \equiv \triangle P O Q(\mathrm{SAS})\)

Hence, \(A B=P Q\) (matching sides of congruent triangles)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-15(1)}
\end{center}

\section*{Chords and angles at the centre}
\begin{itemize}
  \item Chords of equal length subtend equal angles at the centre of a circle.
  \item Conversely, chords subtending equal angles at the centre of a circle have equal length.
\end{itemize}

\section*{The midpoint of a chord}
Three theorems about the midpoint of a chord are stated below. The proofs are dealt with in question 6 of Exercise 13C.

Theorem: The interval joining the midpoint of a chord to the centre of a circle is perpendicular to the chord, and bisects the angle at the centre subtended by the chord.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-15(2)}
\end{center}

Theorem: The perpendicular from the centre of a circle to a chord bisects the chord, and bisects the angle at the centre subtended by the chord.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-15}
\end{center}

Theorem: The bisector of the angle at the centre of a circle subtended by a chord bisects the chord, and is perpendicular to it.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-15(3)}
\end{center}

\section*{Chords and calculations}
The circle theorems stated above can be used in conjuction with Pythagoras' theorem and trigonometry to calculate lengths and angles.

\section*{Example 5}
A chord of length \(12 \mathrm{~cm}\) is drawn in a circle of radius \(8 \mathrm{~cm}\).

a How far is the chord from the centre (that is, the perpendicular distance from the centre to the chord)?

b What angle, correct to one decimal place, does the chord subtend at the centre?

\section*{Solution}
a Draw the perpendicular \(O M\) from \(O\) to the chord. By a midpoint of the chord theorem, \(M\) is the midpoint of \(A B\), so \(A M=6\).

Hence, \(O M^{2}=8^{2}-6^{2} \quad\) (Pythagoras' theorem)

\[
\begin{aligned}
O M & =\sqrt{28} \\
& =2 \sqrt{7} \mathrm{~cm}
\end{aligned}
\]

b Let \(\theta=\angle A O M\)

Then, \(\sin \theta=\frac{6}{8}\)

\[
\theta \approx 48.59^{\circ}
\]

So, \(\angle A O B=2 \theta \approx 97.2^{\circ}\)

\section*{The midpoint of a chord}
\begin{itemize}
  \item The interval joining the midpoint of a chord to the centre of a circle is perpendicular to the chord, and bisects the angle at the centre subtended by the chord.
  \item The perpendicular from the centre of a circle to a chord bisects the chord, and bisects the angle at the centre subtended by the chord.
  \item The bisector of the angle at the centre of a circle subtended by a chord bisects the chord, and is perpendicular to it.
\end{itemize}

\section*{Finding the centre of a circle}
Suppose that we have a circle. How do we find its centre?

The first midpoint-of-a-chord theorem above tells us that the centre lies on the perpendicular bisector of every chord. Thus, if we draw two chords that are not parallel, and construct their perpendicular bisectors, the point of intersection of the bisectors will be the centre of the circle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-16(1)}
\end{center}

\section*{The circumcircle of a triangle}
Here is an important fact about circles.

Suppose that three points \(A, B\) and \(C\) form a triangle, meaning that they are not collinear. Then there is a circle passing through all three points. The circle is called the circumcircle of \(\triangle A B C\), and its centre is called the circumcentre of the triangle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-16}
\end{center}

Here is a simple construction of the circumcentre and circumcircle.

Construct the perpendicular bisectors of two sides \(A B\) and \(B C\), and let them meet at \(O\). Then \(O\) is the circumcentre of \(\triangle A B C\), and we can use it to draw the circumcircle through \(A, B\) and \(C\).

Here is the theorem that justifies all the previous remarks.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-17(1)}
\end{center}

Theorem: The intersection of the perpendicular bisectors of two sides of a triangle is the centre of a circle passing through all three verticles.

Proof: \(\quad\) Let \(A B C\) be a triangle.

Let \(M\) be the midpoint of \(A B\), and let \(N\) be the midpoint of \(B C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-17}
\end{center}

Let the perpendicular bisectors of \(A B\) and \(B C\) meet at \(O\), and join \(A O, B O\) and \(C O\).

Then \(\triangle A O M \equiv \triangle B O M(\mathrm{SAS})\)

so \(\quad A O=B O\) (matching sides of congruent triangles),

and \(\triangle C O N \equiv \triangle B O N(\mathrm{SAS})\)

so

\[
\begin{aligned}
& C O=B O \text { (matching sides of congruent triangles), } \\
& A O=B O=C O
\end{aligned}
\]

Hence, \(O\) is equidistant from \(A, B\) and \(C\), so the circle with centre \(O\) and radius \(A O\) passes through \(A, B\) and \(C\).

\section*{The centre of a circle and circumcentre of a triangle}
\begin{itemize}
  \item To find the centre of a given circle, construct the perpendicular bisectors of two nonparallel chords, and take their point of intersection.
  \item To find the circumcentre of a given triangle, construct the perpendicular bisectors of two sides, and take their point of intersection.
\end{itemize}

\section*{Exercise 13C}
Note: Points labelled \(O\) in this exercise are always centres of circles.

1 a Draw a large circle (and ignore the fact that you may be able to see the mark that the compasses made at the centre). Draw two non-parallel chords \(A B\) and \(P Q\), then construct their perpendicular bisectors. The point where the bisectors intersect is the centre of the circle.

b Draw a large triangle \(A B C\).

i Construct the perpendicular bisectors of two sides, and let them intersect at \(O\). Construct the circle with circumcentre \(O\) passing through all three vertices of the triangle.

ii Construct the perpendicular bisector of the third side. It should also pass through \(O\).

2 Find the exact value of \(x\), as a surd if necessary. Then use trigonometry to find the value of \(\theta\), correct to one decimal place.

\(\mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18(7)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18(5)}
\end{center}

3 a A chord subtends an angle of \(90^{\circ}\) at the centre of a circle of radius \(12 \mathrm{~cm}\).

i How long is the chord?

ii How far is the midpoint of the chord from the centre?

b In a circle of radius \(20 \mathrm{~cm}\), the midpoint of a chord is \(16 \mathrm{~cm}\) from the centre.

i How long is the chord?

ii What angle does the chord subtend at the centre, correct to one decimal place?

c In a circle of radius \(10 \mathrm{~cm}\), a chord has length \(16 \mathrm{~cm}\).

i What is the perpendicular distance from the chord to the centre?

ii What angle does the chord subtend at the centre, correct to one decimal place?

4 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18(1)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18(4)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-18(3)}
\end{center}

f
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-18(8)}

5 Let two circles of radius 1 and centres \(O\) and \(P\) each pass through the centre of the other, and intersect at \(F\) and \(G\).

Let \(F G\) meet \(O P\) at \(M\).

a Find \(\angle F P O, \angle F G O\) and \(\angle F M O\).

b Find the length of the common chord \(F G\).

6 This question leads you through the proofs of the three theorems in the text about the midpoint of a chord.

a Prove that: The line joining the midpoint of a chord to the centre is perpendicular to the chord, and bisects the angle at the centre subtended by the chord.

Let \(M\) be the midpoint of the chord \(A B\).

i Prove that \(\triangle A O M \equiv \triangle B O M\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-19(3)}
\end{center}

ii Hence, prove that \(O M \perp A B\) and that \(O M\) bisects \(\angle A O B\).

b Prove that: The perpendicular from the centre to a chord bisects the chord, and bisects the angle at the centre subtended by the chord.

Let \(M\) be the foot of the perpendicular from \(O\) to chord \(A B\).

i Prove that \(\triangle A O M \equiv \triangle B O M\).

ii Hence, prove that \(M\) is the midpoint of \(A B\) and that \(O M\) bisects \(\angle A O B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-19(1)}
\end{center}

c Prove that: The bisector of the angle at the centre subtended by a chord bisects the chord, and is perpendicular to it.

Let the bisector of \(\angle A O B\) meet the chord \(A B\) at \(M\).

i Prove that \(\triangle A O M \equiv \triangle B O M\).

ii Hence, prove that \(M\) is the midpoint of \(A B\) and that \(O M \perp A B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-19(2)}
\end{center}

\(\begin{array}{ll}7 & \mathbf{a} \\ & \end{array}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-19}
\end{center}

i Prove that \(\triangle A O P \equiv \triangle A O Q\).

ii Prove that \(A P=A Q\).

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-19(6)}
\end{center}

i Prove that \(\triangle P S T \equiv \triangle Q S T\).

ii Prove that \(\angle P=\angle Q\).

iii Prove that \(\angle P+\angle Q=180^{\circ}\).

iv Prove that \(S T\) is a diameter. b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-19(5)}
\end{center}

i Prove that \(\angle F S O=\angle T F S\).

ii Prove that \(F T \| O S\).

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-19(4)}
\end{center}

i Prove that \(\angle O S T=\angle O T S\).

ii Prove that \(\angle O R U=\angle O U R\).

iii Prove that \(\triangle O R T \equiv \triangle O U S\).

iv Prove that \(R S=T U\).

8 Prove that: When two circles intersect, the line joining their centres is the perpendicular bisector of their common chord.

Let two circles with centres \(O\) and \(P\) intersect at \(F\) and \(G\).

Let \(O P\) meet the common chord \(F G\) at \(M\).

a Prove that \(\triangle F O P \equiv \triangle G O P\).

b Prove that \(\triangle F O M \equiv \triangle G O M\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-20(2)}
\end{center}

c Hence, prove that \(F M=G M\) and \(O P \perp F G\).

9 Let \(A B\) be an interval with midpoint \(M\), and let \(P\) be a point in the plane not on \(A B\).

a Prove that if \(P\) is equidistant from \(A\) and \(B\), then \(P\) lies on the perpendicular bisector of \(A B\).

b Conversely, prove that if \(P\) lies on the perpendicular bisector of \(A B\), then \(P\) is equidistant from \(A\) and \(B\).

c Use parts \(\mathbf{a}\) and \(\mathbf{b}\) to prove that a circle has only one centre.

d For these questions you will need to think in three dimensions.

i A circle is drawn on a piece of paper that lies flat on the table. Is there any other point in three-dimensional space, other than the centre of the circle, that is equidistant from all the points on the circle?

ii What geometrical object is formed by taking, in three dimensions, the set of all points that are a fixed distance from a given point?

iii What geometrical object is formed by taking, in three dimensions, the set of all points that are a fixed distance from a given line?

iv What geometrical object is formed by taking, in three dimensions, the set of all points that are a fixed distance from a given interval?

v What geometrical object is formed by taking, in three dimensions, the set of all points that are equidistant from the endpoints of an interval?

vi How could you find the centre of a sphere?

\section*{13 Tangents and radij}
The diagrams show that a line can intersect a circle at two points, one point or no points.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-20(1)}
\end{center}

two points

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-20(3)}
\end{center}

one point

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-20}
\end{center}

no points

\begin{itemize}
  \item A line that intersects a circle at two points is called a secant, because it cuts the circle into two pieces. (The word secant comes from Latin and means 'cutting'.)
  \item A line that intersects a circle at just one point is called a tangent. It touches the circle at that point of contact, but does not pass inside it. (The word tangent also comes from Latin and means 'touching'.)
\end{itemize}

\section*{Constructing a tangent}
In the diagram, \(O T\) is a radius of a circle. The line \(P T Q\) is perpendicular to the radius \(O T\).

Could \(P Q\) intersect the circle at a second point (other than \(T\) )? The symmetry of the diagram about the line \(O T\) suggests that it cannot, and here is the proof.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-21(2)}
\end{center}

Theorem: The line through a point on a circle perpendicular to the radius at that point is the tangent at that point.

Proof: \(\quad\) Let \(O T\) be a radius, and let \(P T Q\) be perpendicular to \(O T\). Let \(X\) be a point other than \(T\) on \(P T Q\).

Then \(O X^{2}=O T^{2}+T X^{2} \quad\) (Pythagoras' theorem) \(O X^{2}>O T^{2}\) since \(T X\) is non-zero,

so \(\quad O X>O T\), and \(O T\) is the radius of the circle

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-21}
and thus \(X\) lies outside the circle.

Hence, \(P T Q\) intersects the circle at only the one point, \(T\), and so \(P T Q\) is a tangent to the circle.

\section*{Example 6}
In the diagram, \(T P\) is a tangent to the circle with centre \(O\).

a Find \(O P\) and \(M P\).

b Find \(\angle T O P\), correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-21(1)}
\end{center}

\section*{Solution}
a We know that \(O T \perp T P\) (radius and tangent),

\[
\begin{aligned}
\text { so } O P^{2} & =10^{2}+15^{2} \quad \text { (Pythagoras' theorem) } \\
& =325
\end{aligned}
\]

\[
O P=5 \sqrt{13}
\]

Hence, \(M P=5 \sqrt{13}-10\)

b Let \(\theta=\angle T O P\)

Then, \(\tan \theta=\frac{15}{10}\)

\[
\theta \approx 56.3^{\circ}
\]

\section*{Tangents from an external point}
Let \(P\) be a point outside a circle. The diagram shows how different lines through \(P\) intersect the circle at two, one or no points. You can see that there are clearly exactly two tangents to the circle from \(P\).

We now prove that these two tangents from the point \(P\) to the circle have equal length.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-22(2)}
\end{center}

Theorem: The tangents to a circle from a point outside have equal length.

Proof: \(\quad\) Let \(P\) be a point outside the circle with centre \(O\).

Let the tangents from \(P\) touch the circle at \(S\) and \(T\).

In the triangles \(O P S\) and \(O P T\),

\[
\begin{aligned}
& O P=O P(\text { common }) \\
& O S=O T(\text { radii })
\end{aligned}
\]

\[
\angle P S O=\angle P T O=90^{\circ}(\text { radius and tangent })
\]

so \(\quad \triangle O P S \equiv \triangle O P T(\mathrm{RHS})\)

Hence, \(P S=P T\) (matching sides of congruent triangles)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-22}
\end{center}

Alternative proof: \(\triangle P S T\) and \(\triangle P T O\) are right-angles

\[
\text { Therefore, } \begin{aligned}
P S^{2} & =P O^{2}-O S^{2}(\text { Pythagoras' theorem }) \\
& =P O^{2}-O T^{2}(\text { radii of a circle }) \\
& =P T^{2}(\text { Pythagoras' theorem })
\end{aligned}
\]

\section*{Example 7}
The intervals \(P S\) and \(P T\) are tangents.

Find \(\theta\) and \(x\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-22(1)}
\end{center}

\section*{Solution}
First, \(x=7\) (tangents from an external point)

Hence, \(\angle T=\theta\) (base angles of isosceles \(\triangle P S T\) )

so \(\theta=70^{\circ}\) (angle sum of \(\left.\triangle P S T\right)\)

\section*{Common tangents and touching circles}
The five diagrams below show all the ways in which two circles of different radii can intersect. Start with the smaller circle inside the larger, and move it slowly to the right. Notice that the two circles can intersect at two, one or no points.
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-23(3)}

The various lines are all the common tangents to the two circles. There are \(0,1,2,3\) and 4 common tangents in the five successive diagrams.

In the second and fourth diagrams, the two circles touch each other, and they have a common tangent at the point of contact.

\section*{Tangents to a circle}
\section*{Tangent and radius}
\begin{itemize}
  \item The line through a point on a circle perpendicular to the radius at that point is the tangent at that point.
\end{itemize}

\section*{Tangents from outside the circle}
\begin{itemize}
  \item The tangents to a circle from a point outside have equal length.
\end{itemize}

\section*{Exercise 13D}
Note: Points labelled \(O\) in this exercise are always centres of circles.

1 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons. In each diagram, a tangent is drawn at \(T\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-23(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-23(2)}
\end{center}

c

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-23}
d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(11)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(9)}
\end{center}

j

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(6)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(4)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(3)}
\end{center}

\(\mathbf{k}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(13)}
\end{center}

\(\mathbf{f}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(7)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(12)}
\end{center}

1

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(1)}
\end{center}

2 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), and the values of \(x, y\) and \(z\). In the diagrams, tangents are drawn to the circle at \(S, T\) and \(U\).

\(\mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(8)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(10)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(14)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24(5)}
\end{center}

f \(P\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-24}
\end{center}

3 Find the values of \(x, y\) and \(\theta\) correct to two decimal places, where tangents are drawn at \(S\) and \(T\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-25(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-25(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-25(5)}
\end{center}

4 Prove that: The tangents at the endpoints of a diameter are parallel.

Let \(P A Q\) and \(U B V\) be the tangents at the endpoints of a diameter \(A O B\). Prove that \(P Q \| U V\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-25(6)}
\end{center}

5 The tangents at the four points \(P, Q, R\) and \(S\) on a circle form a quadrilateral \(A B C D\). Prove that \(A B+C D=A D+B C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-25(4)}
\end{center}

6 This question describes the method of construction of tangents to a circle from an external point \(P\).

a Draw a circle with centre \(O\) and choose a point, \(P\), outside the circle. Let \(M\) be the midpoint of \(O P\), and hence draw the circle with diameter \(O P\). Let the circles intersect at \(S\) and \(T\), and join \(P S\) and \(P T\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-25}
\end{center}

b Prove that \(P S\) and \(P T\) are tangents to the original circle.

c Deduce that \(P S=P T\).

7 Let \(P S\) and \(P T\) be the two tangents to a circle with centre \(O\) from a point, \(P\), outside the circle.

a i Prove that \(\triangle P S O \equiv \triangle P T O\).

ii Hence, prove that the tangents have equal length, and that \(O P\) bisects the angle between the tangents and bisects the angle between the radii at \(O S\) and \(O T\).

b Join the chord \(S T\) and let it meet \(P O\) at \(M\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-25(2)}
\end{center}

i Prove that \(\triangle S P M \equiv \triangle T P M\).

ii Hence, prove that \(O P\) is the perpendicular bisector of \(S T\).

8 The circle in the diagram is called the incircle of triangle \(A B C\). It touches the three sides of \(\triangle A B C\) at \(P, Q\) and \(R\). Prove that: Area of triangle \(=\frac{1}{2} \times(\) perimeter of triangle \() \times(\) radius of circle \()\) You will need to join the radii \(O P, O Q\), and \(O R\) and the intervals \(O A, O B\) and \(O C\), where \(O\) is the centre of the

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-26(3)}
incircle.

\(9 \mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-26(2)}
\end{center}

i Prove that \(\alpha=30^{\circ}\).

ii Find \(\beta\). b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-26}
\end{center}

i Prove that \(\sin \alpha=\frac{1}{2}\).

ii Prove that \(\beta=60^{\circ}\).

10 Prove that: When two circles touch, their centres and their point of contact are collinear.

a Let two circles with centres \(O\) and \(P\) touch externally at \(T\). Let \(A T B\) be the common tangent at \(T\).

i Find \(\angle A T O\) and \(\angle A T P\).

ii Hence, prove that \(O, T\) and \(P\) are collinear.

b Draw a diagram of two circles touching internally, and

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-26(1)}
prove the theorem in this case.

11 a Each tangent, \(F R\) and \(G S\), in the diagram is called a direct common tangent because the two circles lie on the same side of the tangent. Produce the two tangents to meet at \(M\).

i Prove that \(M F=M G\) and \(M R=M S\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-26(4)}
\end{center}

ii Hence, prove that \(F R=G S\).

b Draw a diagram showing indirect common tangents, and prove that they also have equal length. (Note: Indirect common tangents cross over, and intersect between the two circles.)

12 Let \(A B\) be a direct common tangent of two circles touching externally. Let the common tangent at the point of contact, \(T\), meet \(A B\) at \(M\).

a Prove that \(M A=M B=M T\).

b Hence, deduce that \(\angle A T B\) is a right angle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-26(5)}
\end{center}

\section*{The alternate segment theorem}
Consider the secant \(X Y\) that intersects a circle at points \(A\) and \(Q\). Consider also points \(P\) and \(B\) and the angles subtended at \(P\) and \(Q\) by the \(\operatorname{arc} A B\).
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-27(3)}

As you can see from the images above, as \(Q\) approaches \(A\) along the circumference, \(\angle X Q B\) and \(\angle P\) remain equal. It therefore seems reasonable to suppose that as \(Q\) coincides with \(A\), and \(X Y\) becomes tangent to the circle at \(A, \angle X A B\) and \(\angle P\) are equal.

This is in fact the case, and is the alternate segment theorem.

The alternate segment ('alternate' here simply means 'other') is the segment of the circle on the other side of the chord \(A B\) from \(\angle X A B\). The angle \(\angle P\) is an angle in the alternate segment.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-27(1)}
\end{center}

Theorem: The angle between a tangent and a chord is equal to any angle in the alternate segment.

Proof: \(\quad\) Let \(A B\) be a chord of a circle and let \(X A Y\) be a tangent at \(A\). Let \(P\) be a point on the circle on the other side of the chord \(A B\) from \(\angle X A B\). Let \(\angle P=\theta\).

We must prove that \(\angle X A B=\angle P\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-27}
\end{center}

Draw the diameter \(A O N\), and join \(B N\).

Then \(\angle N=\theta(\) angles on the same \(\operatorname{arc} A B)\)

and \(\angle N B A=90^{\circ}\) (angle in the semicircle \(N B A\) )

and \(\angle N A X=90^{\circ}\) (radius and tangent)

Hence, \(\angle N A B=90^{\circ}-\theta(\) angle sum of \(\triangle N A B)\)

so \(\angle X A B=\theta\) (adjacent angles in a right angle)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-27(2)}
\end{center}

Note: This proof is only valid when \(\angle X A B\) is acute. In the exercises we will prove the result when \(\angle X A B\) is obtuse.

\section*{Example 8}
Find \(\alpha, \beta, \gamma\) and \(\theta\) in the figure shown to the right.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-28(3)}
\end{center}

\section*{Solution}
\(\alpha=70^{\circ}\) (alternate segment theorem)

\(\beta=70^{\circ}\) (alternate angles, \(D E \| A C\) )

\(\gamma=70^{\circ}\) (alternate angles, \(D E \| A C\) )

\(\theta=40^{\circ}(\) angles in a straight angle at \(B)\)

\section*{The alternate segment theorem}
The angle between a tangent and a chord is equal to any angle in the alternate segment.

\section*{Exercise 13E}
Note: Points labelled \(O\) in this exercise are always centres of circles.

1 Draw a large circle and a chord \(A B\). At one end of the chord, draw the tangent to the circle.

a Mark one of the angles \(\theta\) between the tangent and the chord \(A B\), then draw any angle in the alternate segment. How are these two angles related?

b Mark the angle \(\alpha\) between the tangent and the chord \(A C\). Which angle is equal to \(\alpha\) ?

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-28}
\end{center}

2 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons. In each diagram, a tangent is drawn at \(T\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-28(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-28(1)}
\end{center}

c

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-28(2)}
d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(9)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(8)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29}
\end{center}

3 Find the values of \(\alpha, \beta, \gamma\) and \(\theta\), giving reasons. In the diagrams, tangents are drawn at \(S, T\) and \(U\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(1)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(5)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(7)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(10)}
\end{center}

4 Here is a different proof of the alternate segment theorem.

Let \(A B\) be a chord of a circle, and let \(S A T\) be the tangent at \(A\).

Let \(\theta=\angle B A T\) be an acute angle. We must prove that \(\angle A P B=\theta\).

a Join the radii \(O A\) and \(O B\). What is the size of \(\angle B A O\) ?

b What is the size of \(\angle A O B\) ?

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-29(2)}
\end{center}

c Hence, prove that \(\angle A P B=\theta\).

5 Show that the alternate segment theorem holds when the angle between the tangent and the chord is obtuse.

Let \(\angle P=\alpha\). We must prove \(\angle S A B=\alpha\).

a Construct diameter \(A N\) and chord \(N B\). Show that \(\angle A N B=180^{\circ}-\alpha\).
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-29(4)}
b Show that \(\angle N A B=\alpha-90^{\circ}\).

c Using \(\angle S A B=\angle S A N+\angle N A B\), show that \(\angle S A B=\alpha\).

d Use the technique used in question 4 to prove the alternate segment theorem for an obtuse angle.

6 Choose \(T\) on the smaller circle. Let \(F T G\) be the tangent at \(T\) and construct lines \(K A T\) and \(T B L\) as shown in the diagram. Let \(\angle L K T=\theta\).

a Prove that \(\angle G T A=\theta\).

b Hence, prove that \(L K \| F G\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-30(2)}
\end{center}

7 The two circles in the diagram touch externally at \(T\), with common tangent \(A T B\) at the point of contact, and FTP and \(G T Q\) are straight lines.

a Let \(\angle F=\theta\). Prove that \(\angle G T B=\theta\) and \(\angle Q T A=\theta\).

b Hence, prove that \(F G \| Q P\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-30(3)}
\end{center}

8 The two circles in the diagram touch externally at \(T\), with common tangent \(A T B\) at the point of contact. Suppose \(P, T\) and \(F\) are collinear and \(G F \| Q P\).

a Let \(\theta=\angle F\). Prove that \(\angle P=\theta\).

b Hence, prove that the points \(G, T\) and \(Q\) are collinear.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-30}
\end{center}

\section*{13 similarity and circles}
\section*{Intersecting chords}
Take a point, \(M\), inside a circle, and draw two chords, \(A M B\) and \(Q M P\), through \(M\). Each chord is thus divided into two subintervals called intercepts. In the following, we shall prove that:

\[
A M \times B M=P M \times Q M
\]

This is a very interesting and useful result called the intersecting chord theorem.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-30(1)}
\end{center}

Theorem:

When two chords of a circle intersect, the product of the intercepts on one chord equals the product of the intercepts on the other chord.

Proof: Draw the intervals \(A P\) and \(B Q\) to make two triangles \(A M P\) and \(Q M B\).

\(\triangle A M P\) is similar to \(\triangle Q M B\) (AAA).

Hence,

\(\frac{A M}{Q M}=\frac{P M}{B M}\) (matching sides of similar triangles)

so, \(A M \times B M=P M \times Q M\).

Note: If we have a family of chords passing through a point, we can apply the theorem to see that \(a b=c d=e f\).
\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-31}

\section*{Intercepts}
A point \(M\) on an interval \(A B\) divides that interval into two subintervals \(A M\) and \(M B\), called intercepts.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-31(2)}
\end{center}

For the next two theorems, we will need to apply this definition to the situation where the dividing point \(M\) is still on the line \(A B\), but is outside the interval \(A B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-31(1)}
\end{center}

The intercepts are still \(A M\) and \(B M\). Everything works in exactly the same way provided that both intercepts are measured from \(M\).

\section*{Secants from an external point}
Now take a point, \(M\), outside a circle, and draw two secants, \(M B A\) and \(M Q P\), to the circle from \(M\). Provided that we continue to take our lengths from \(M\) to the circle, the statement of the result is the same. That is:

\[
A M \times B M=P M \times Q M
\]

Theorem: When two secants intersect outside a circle, the product of the intercepts on one secant equals the product of the intercepts on the other secant.

The proof by similarity is practically the same as when \(M\) is inside the circle,

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-31(3)}
and we will address this in question 2 of Exercise 13F.

\section*{Example 9}
Find \(x\) in each diagram.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-32(1)}
\end{center}

b \(M\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-32(2)}
\end{center}

\section*{Solution}
a Using intersecting chords:

\[
\begin{aligned}
3 \times x & =6 \times 4 \\
x & =8
\end{aligned}
\]

b Using secants from an external point:

\[
\begin{aligned}
8 \times(8+x) & =6 \times(6+10) \\
8(8+x) & =96 \\
8+x & =12 \\
x & =4
\end{aligned}
\]

\section*{Tangent and secant from an external point}
As the point \(Q\) moves towards \(T\), the line \(M Q P\) becomes the tangent at \(T\). Thus, the previous product \(P M \times Q M\) has become the square \(T M^{2}\). That is:

\[
A M \times B M=T M^{2}
\]

We therefore have a new theorem. For completeness, we give another proof.

Theorem: When a secant and a tangent to a circle intersect, the product of

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-32(3)}
the intercepts on the secant equals the square of the tangent.

That is, \(A M \times B M=T M^{2}\)

Proof: \(\quad\) Let \(M\) be a point external to a circle.

Let \(T M\) be a tangent from \(M\). Suppose a secant from \(M\) cuts the circle at \(A\) and \(B\).

Draw the intervals \(A T\) and \(B T\), and look at the two triangles \(A M T\) and \(T M B\).

\[
\begin{aligned}
& \angle A M T=\angle T M B(\text { common angle }) \\
& \angle M A T=\angle M T B(\text { alternate segment theorem })
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-32}
\end{center}

so \(\triangle A M T\) is similar to \(\triangle T M B\) (AAA).

Hence, \(\frac{A M}{T M}=\frac{T M}{B M}\) (matching sides of similar triangles)

so \(A M \times B M=T M^{2}\)

\section*{Example 10}
Find \(x\) in each diagram, given that \(M T\) is a tangent to the circle.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-33(7)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-33}
\end{center}

\section*{Solution}
We use the tangent and secant theorem in each part.
a \(x^{2}=5 \times(5+15)\)
b

\[
x^{2}=100
\]

\[
x=10 \text { (since } x \text { is positive) }
\]

b \(\quad x \times(x+5)=6^{2}\)
\(x^{2} \times 5 x-36=0\)
\((x+9)(x-4)=0\)
\(x=4\) (since \(x\) is positive)

\section*{Chords, secants and tangents}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-33(6)}
\end{center}

\(A M \times B M=P M \times Q M\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-33(5)}
\end{center}

\(A M \times B M=P M \times Q M\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-33(2)}
\end{center}

\(A M \times B M=T M^{2}\)

\section*{Exercise 13F}
Note: Points labelled \(O\) in this exercise are always centres of circles.

1 In each diagram, find the value of \(x\), giving reasons. Tangents are drawn at the point \(T\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-33(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-33(4)}
\end{center}

c

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-33(3)}
d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(4)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(7)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(2)}
\end{center}

\(\mathbf{k}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(5)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(1)}
\end{center}

I

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(6)}
\end{center}

2 Let secants from a point, \(M\), external to the circle cut the circle at points \(A\) and \(B\) and \(P\) and \(Q\), as shown in the diagram. Prove that \(A M \times B M=P M \times Q M\).

This is the proof of the theorem stated on page 410 .

3 Prove the result of Question 2 by drawing a tangent \(M T\) and using the 'tangent and secant' theorem.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(8)}
\end{center}

4 Let \(A O B\) be a diameter of a circle, and let \(G H\) be a chord perpendicular to \(A B\), meeting \(A B\) at \(M\).

a Why is \(M\) the midpoint of \(G H\) ?

b Let \(g=G M, a=A M\) and \(b=B M\). Prove that \(g^{2}=a b\).

c Explain why the radius of the circle is \(\frac{a+b}{2}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-34(3)}
\end{center}

d Prove that \(\sqrt{a b} \leq \frac{a+b}{2}\).

This is the well known Arithmetic mean-Geometric mean inequality.

5 Let \(P\) be a point on the common secant \(A B\) of two intersecting circles. Let \(P S\) and \(P T\) be tangents from \(P\), one to each circle. Prove that \(P S=P T\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(8)}
\end{center}

6 In the diagram, \(M S\) and \(M T\) are tangents from an external point, \(M\).

a Prove that \(\triangle M S A\) is similar to \(\triangle M B S\).

b Hence, prove that \(\frac{a}{x}=\frac{t}{m}\).

c Similarly, prove that \(\frac{y}{b}=\frac{t}{m}\).

d Hence, prove that \(a b=x y\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35}
\end{center}

\section*{Review exercise}
1 Find the values of the pronumerals.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(6)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(5)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(3)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(2)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(4)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(9)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(7)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-35(10)}
\end{center}

b

2 Find the values of the pronumerals.

a

3 Find the values of the pronumerals.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-36(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-36(4)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-36(3)}
\end{center}

\(4 A B C D\) is a cyclic quadrilateral. Its diagonals \(A C\) and \(B D\) intersect at \(P\). Prove that \(\triangle A P D\) is similar to \(\triangle B P C\).

5 The quadrilateral \(A B C D\) has its vertices on a circle with centre \(O\). The diagonal \(A B\) is a diameter of the circle and \(A C=B D\). Prove that \(A D=B C\).

\(6 A B C D\) is a cyclic quadrilateral with \(A D\) parallel to \(B C\). The diagonals \(A C\) and \(B D\) intersect at \(P\). Prove that \(\angle A P B=2 \angle A C B\).

\(7 \quad A B C D\) is a cyclic quadrilateral. Chord \(A B\) is produced and a point \(E\) is marked on the line \(A B\) so that \(B\) is between \(A\) and \(E\). Prove that \(\angle E B C=\angle A D C\).

\(8 P Q R S\) is a cyclic quadrilateral. The diagonal \(P R\) bisects both \(\angle S P Q\) and \(\angle S R Q\). Prove that \(\angle P Q R\) is a right angle.

9 In the diagram opposite, the two circles intersect at \(B\) and \(E\).

Prove that \(A F\) is parallel to \(C D\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-36(1)}
\end{center}

10 Two circles intersect at \(T\) and \(V\). The intervals \(P T Q\) and \(R T S\) are drawn as shown. Prove that \(\angle P V R=\angle Q V S\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-36(5)}
\end{center}

11 In the figure, \(A O B\) is the diameter of the circle \(A B C\) with centre \(O\). The point \(Q\) is the centre of another circle that passes through the points \(A, O\) and \(C\), and \(Q X \perp A C\).

a Prove that \(\angle A Q X=2 \angle A B C\).

b Show that \(A B^{2}=B C^{2}+4\left(A Q^{2}-X Q^{2}\right)\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-36}
\end{center}

\section*{Challenge exercise}
1 Here is a form of the sine rule that shows its connection with the circumcircle of a triangle.

In any triangle \(A B C, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R\),

where \(R\) is the radius of the circumcircle of triangle \(A B C\). Assume that \(A\) is acute.

Draw the circumcircle of \(\triangle A B C\), and let \(O\) be the circumcentre.

a Hence, prove that \(\frac{a}{\sin A}=2 R\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-37(1)}
\end{center}

b Prove the result when \(\angle A\) is obtuse.

2 Prove that: In any triangle \(A B C\), the bisectors of the vertex angles are concurrent, and the resulting incentre is the centre of a circle that touches all sides of the triangle.

Let the angle bisectors of \(\angle B\) and \(\angle C\) meet at \(I\), and draw \(I A\). Draw the perpendiculars \(I P, I Q\) and \(I R\) to the sides \(B C, C A\) and \(A B\), respectively.

a Use congruence to prove that \(I R=I P\), and that \(I P=I Q\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-37(3)}
\end{center}

b Use congruence to prove that \(I A\) bisects \(\angle A\).

c Why does the circle with centre \(I\) and radius \(I R\) touch all three sides of the triangle?

3 Prove that: In any triangle \(A B C\), the altitudes are concurrent (their intersection is called the orthocentre of the triangle).

Let the altitudes \(A K\) and \(B L\) meet at \(H\). Draw \(C H\) and produce it to meet \(A B\) at \(M\).

a Prove that \(C, K, H\) and \(L\) are concyclic.

b Let \(\angle A C M=\theta\). Prove that \(\angle A K L=\theta\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-37}
\end{center}

c Prove that \(B, K, L\) and \(A\) are concyclic, and hence prove that \(\angle A B L=\theta\).

d Hence, prove that \(C M\) is an altitude of the triangle. That is, we have proved that the three altitudes are concurrent.

4 Euler discovered a wonderful theorem that the Greeks had missed:

The orthocentre, the centroid and the circumcentre of a triangle are collinear, with the centroid dividing the interval joining the orthocentre and circumcentre in the ratio \(2: 1\).

(This line is called the Euler line.)

Note: The centroid is the point of intersection of the medians of a triangle, which are the lines drawn from any vertex of a triangle to the midpoint of the opposite side.

Let \(O\) and \(G\) be the circumcentre and centroid, respectively, of \(\triangle A B C\). Draw \(O G\) and produce it to a point, \(M\), such that

\includegraphics[max width=\textwidth, center]{2024_02_22_e44df152b05c6be06c4cg-37(2)}
\(O G: G M=1: 2\).
a Let \(F\) be the midpoint of \(B C\). Use the fact that the centroid, \(G\), divides the median \(A F\) in the ratio \(2: 1\) to prove that \(\triangle G O F\) is similar to \(\triangle G M A\).

b Hence, prove that \(M\) lies on the altitude from \(A\).

c Show that point \(M\) is the point \(H\) constructed in Question 3.

5 Prove the following converse of the cyclic quadilateral theorem:

If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Let the opposite angles of the quadrilateral \(A B C D\) be supplementary. Draw the circle through the points \(A, B\) and \(C\). Let \(A D\), produced if necessary, meet the circle at \(P\), and draw \(P C\).

a Prove that \(\angle P=\angle D\).

b Hence, prove that the points \(P\) and \(D\) coincide.

6 Prove the following converse of the intersecting chords theorem:

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-38}
\end{center}

Suppose that two intervals \(A B\) and \(C D\) intersect at \(M\), and that \(A M \times B M=C M \times D M\). Then the points \(A, B, C\) and \(D\) are concyclic.

Draw the circle through the points \(A, B\) and \(C\). Let \(C D\), produced if necessary, meet the circle at \(P\).

a Prove that \(P M \times C M=A M \times B M\).

b Hence, prove that the points \(P\) and \(D\) coincide.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-38(1)}
\end{center}

7 Take two non-intersecting circles in the plane with centres \(O\) and \(O^{\prime}\). Draw two indirect common tangents \(A A^{\prime}\) and \(B B^{\prime}\), and one direct tangent \(C C^{\prime}\), where \(A, B\) and \(C\) lie on the first circle, and \(A^{\prime}, B^{\prime}\) and \(C^{\prime}\) lie on the second circle. Produce \(A A^{\prime}\) and \(B B^{\prime}\) to meet \(C C^{\prime}\) at \(X\) and \(Y\).

a Prove that \(A A^{\prime}=B B^{\prime}\).

b Prove that \(A A^{\prime}=X Y\).

c Describe what happens when the two circles are touching each other externally.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e44df152b05c6be06c4cg-38(2)}
\end{center}

8 Two circles intersect at \(A\) and \(B\). A straight line passing through \(A\) meets the two circles respectively at \(C\) and \(D\).

a Show that any two triangles \(C B D\) formed in this way are similar.

b Which of these triangles has the larger area?

9 Two circles touch externally at \(P\), and a common tangent touches them at \(A\) and \(B\). Let the common tangent at \(P\) meet \(A B\) at \(C\).

a Show that \(C\) is the midpoint of \(A B\).

b A line passing through \(P\) meets the two circles at \(D\) and \(E\). Draw the tangents to each circle at \(D\) and at \(E\). Show that the tangents are parallel.

10 If \(\triangle A B C\) has side lengths \(a, b\) and \(c\), prove that:

\[
\frac{2 \times(\text { Area of } \triangle A B C)}{a b c}=\frac{\sin C}{c}=\frac{\sin A}{a}=\frac{\sin B}{b}
\]


\end{document}