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\title{Congruent triangles }

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\maketitle
Figures are copied everywhere - a photocopier can produce a copy of a figure, craft workers use copies of figures to create designs or patterns, copies of figures can be obtained by tracing, and so on.

The idea of a 'copy' also arises in geometry and is called 'congruence'.

You have previously met the three kinds of geometrical transformation translation, reflection and rotation. Two figures are called congruent if one is the image of the other when you apply one or more of these transformations.

We will see that two triangles are congruent when the sides and angles of one triangle are equal to the sides and angles of the other.

\section*{12 Congruence of figures in the plane}
If we take the five of spades from each of two identical decks of cards, they look exactly the same. We can move one card and place it on top of the other one so that the pictures on the two cards coincide exactly, as shown below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-02(3)}
\end{center}

Two objects like this are called congruent. The word 'congruent' comes from Latin and means 'in agreement' or 'in harmony'. Here is a more precise definition:

Two plane figures are called congruent if one figure can be moved on top of the other, by a sequence of translations, rotations and reflections, so that they coincide exactly. Thus congruent figures have the same size and the same shape.

For example, the two footprints opposite are congruent because the footprint on the right is a reflection of the footprint on the left.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-02(1)}
\end{center}

\section*{Pairing the parts of congruent figures}
When two figures are congruent, we can pair every part of one figure with the corresponding part of the other by finding a sequence of translations, rotations and reflections which moves one figure exactly on top of the other. For example, the two figures below are congruent. Some of the pairings of points with points, intervals with intervals, and angles with angles are shown in the list below.

Can you see how to transform one figure to the other? There are several ways of doing it. Once you have done this you will see the following pairings. We use the notation \(P^{\prime}\) for the image of the point \(P\) under transformation, as introduced in Chapter 18 of ICE-EM Mathematics Year 7.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-02(2)}
\end{center}

\begin{center}
\begin{tabular}{ll}
\(A \leftrightarrow A^{\prime}\) & \(D \leftrightarrow D^{\prime}\) \\
\(B \leftrightarrow B^{\prime}\) & \(E \leftrightarrow E^{\prime}\) \\
\(C \leftrightarrow C^{\prime}\) & \(A B \leftrightarrow A^{\prime} B^{\prime}\) \\
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-02}
\end{center}

\[
\begin{aligned}
& B C \leftrightarrow B^{\prime} C^{\prime} \\
& \angle E A B \leftrightarrow \angle E^{\prime} A^{\prime} B^{\prime} \\
& B D \leftrightarrow B^{\prime} D^{\prime} \\
& \angle B C D \leftrightarrow \angle B^{\prime} C^{\prime} D^{\prime} \\
& B E \leftrightarrow B^{\prime} E^{\prime} \\
& \angle D B E \leftrightarrow \angle D^{\prime} B^{\prime} E^{\prime}
\end{aligned}
\]

If two figures are congruent, then paired intervals have the same length, paired angles have the same size, and paired regions have the same area.

\section*{Congruent figures}
\begin{itemize}
  \item Two plane figures are called congruent if one figure can be moved on top of the other, by a sequence of translations, rotations and reflections, so that they coincide exactly.
  \item Congruent figures have the same shape and the same size.
  \item When two figures are congruent, we can find a transformation that pairs every part of one figure with the corresponding part of the other, so that:
  \item paired angles have the same size
  \item paired intervals have the same length
  \item paired regions have the same area.
\end{itemize}

\section*{Exercise 12A}
1 List the figures in the collection below that are congruent to each of the figures i, ii and iii.

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(3)}
\end{center}

A

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(9)}
\end{center}

D

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(4)}
\end{center}

G

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(1)}
\end{center}

J

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(11)}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-03(5)}
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-03}

K

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(2)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(10)}
\end{center}

C
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-03(7)}

I

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(6)}
\end{center}

L

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-03(8)}
\end{center}

2 Complete the pairings of vertices, sides and angles of these two congruent figures.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-04(1)}
a \(A \leftrightarrow\)
b \(B \leftrightarrow\)
c \(C \leftrightarrow\)
d \(D \leftrightarrow\)
e \(E \leftrightarrow\)
g \(A C \leftrightarrow\)
h \(B D \leftrightarrow\)
f \(A B \leftrightarrow\)
j \(\angle A B C \leftrightarrow\)
\(\mathbf{k} \angle E A B \leftrightarrow\)
i \(E D \leftrightarrow\)
l \(\angle D A B \leftrightarrow\)

\section*{3 Discussion questions}
a A page of writing is reflected in a mirror. Is the image congruent to the real page?

b Think of two large jacaranda trees that you have seen. Are they congruent?

c Is a model of the Sydney Harbour Bridge congruent to the real bridge?

d Two identical twins stand at attention in army uniform. Are they congruent?

e Think of two cumulus clouds of about the same size. Are they congruent?

f Are two copies of the same photograph congruent?

g What lower-case letter, apart from \(b\) itself, is congruent to \(b\) ?

h What digit, apart from 9 itself, is congruent to 9 ?

\section*{12 
 Congruent triangles}
Most geometrical reasoning about congruence that we are going to do involves only congruent triangles.

\section*{Translations}
Here are two congruent triangles. The sides \(A B\) and \(P Q\) are on the same line.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-04}

The translation that takes \(P\) to \(A\) also takes \(Q\) to \(B\) and \(R\) to \(C\). Thus the two triangles are congruent because \(\triangle A B C\) is the image of \(\triangle P Q R\) under a translation.

\section*{Reflections}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-05(1)}
\end{center}

The triangle \(F G E\) is the reflection of the triangle \(B C A\). Thus the two triangles are congruent.

\section*{Rotations}
Triangle \(U B W\) is a rotation of triangle \(A B C\) about \(B\).

Thus the two triangles are congruent.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-05}
\end{center}

\section*{When are two triangles congruent?}
We can see from the examples above that when two triangles are congruent, paired sides have the same length and paired angles have the same size.

The rest of this chapter will develop four tests for two triangles to be congruent. At this stage, however, we simply note that:

If the vertices of two triangles can be paired up so that paired sides have the same length and paired angles have the same size, then they are congruent.

To demonstrate this, cut out one triangle and move it so that it lies exactly on top of the other triangle. This process involves only translations and rotations, and a reflection if you have to turn the triangle over.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-05(2)}

Thus the two triangles above are congruent. You should be able to image how you would transform one triangle so that it lies on top of the other.

\section*{Writing a congruent statement with vertices in matching order}
When writing symbolic congruence statements, we use the symbol \(\equiv\) for 'is congruent to'.

For example, the triangles in the diagram on the previous page are congruent. This is written as:
\(\triangle A B C \equiv \triangle D E F\)
(Read this as 'triangle \(A B C\) is congruent to triangle \(D E F\). .)

It is vitally important to write the vertices of the two triangles in matching order. In the statement above, we wrote the two triangles as \(\triangle A B C\) and \(\triangle D E F\) because the paired vertices are:

\[
A \leftrightarrow D \quad \text { and } \quad B \leftrightarrow E \quad \text { and } \quad A \leftrightarrow F
\]

This attention to detail is useful when it comes to paired sides. We can read off the paired sides from the congruence statement \(\triangle A B C \equiv \triangle D E F\) in the natural way:

\(A B \leftrightarrow D E \quad\) and \(\quad B C \leftrightarrow E F \quad\) and \(\quad C A \leftrightarrow F D\)

\section*{Congruent triangles}
\begin{itemize}
  \item If the vertices of two triangles can be paired up so that paired angles have equal size and paired sides have equal length, then they are congruent.
  \item When writing a congruence statement, always write the vertices of the two congruent triangles in matching order. We can then read off paired angles and paired sides so that the pairings can be read off in the natural way. For example, the statement \(\triangle A B C \equiv \triangle D E F\) means that:
\end{itemize}

\(\angle A=\angle D, \angle B=\angle E, \angle C=\angle F\),

\(A B=D E, B C=E F, C A=F D\).

\section*{Exercise 12B}
1 Write a congruence statement for each pair of congruent triangles.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-06}
\end{center}

b
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-06(1)}

\(\triangle A B C\) is isosceles with \(C A=C B\).

\(\triangle D E F\) is isosceles with \(F D=F E\)
\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-07}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-07(2)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-07(3)}
\end{center}

\section*{2 Congruent triangles: The SSS and AAS tests}
\section*{The SSS congruence test}
To prove that two triangles are congruent, it is not necessary to prove that they have three pairs of equal sides and three pairs of equal angles. In this section and the next, we will investigate the minimum amount of information needed to establish that two triangles are congruent.

Consider the three diagrams below.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-07(1)}

The three triangles shown are clearly not congruent because the \(3 \mathrm{~cm}\) and \(4 \mathrm{~cm}\) can flap about. This shows that just knowing that two pairs of sides are equal is not enough information to establish congruence. Knowing that three pairs of sides are equal, however, is enough to establish congruence, as the discussion below will demonstrate.

This test for congruence is called the SSS congruence test. The initials SSS stand for Side, Side, Side.

\section*{SSS congruence test}
If the three sides of one triangle are equal to the three sides of another, then the two triangles are congruent.

For example, in the diagram below, \(\triangle A B C \equiv \triangle P Q R\) (SSS).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-08}

Thus the SSS congruence test tells us that the angles of any triangle are determined by the three sides.

We can now conclude from the congruence that:

\[
\begin{aligned}
& \angle A=\angle P \\
& \angle C=\angle R \\
& \angle B=\angle Q
\end{aligned}
\]

You should always write the initials of the test used to demonstrate the congruence after the congruence statement.

\section*{Example 1}
Write down a statement that the two triangles below are congruent, giving the appropriate congruence test as a reason.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-08(1)}

\section*{Solution}
\(\Delta K L M \equiv \Delta X Y Z(\mathrm{SSS})\)

We review the construction of triangles with given side lengths.

\section*{Constructing a triangle given three sides}
We will construct a triangle with three given side lengths. Here is an example using the side lengths \(12 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(5 \mathrm{~cm}\).

Step 1: Draw an interval \(A B\) of length \(12 \mathrm{~cm}\).

Step 2: Draw a circle with centre \(A\) and radius \(5 \mathrm{~cm}\).

Step 3: Draw a circle with centre \(B\) and radius \(10 \mathrm{~cm}\) crossing the first circle at \(C\) and \(C^{\prime}\).

Step 4: Join up the triangle \(A B C\) and the triangle \(A B C^{\prime}\).

The two triangles \(A B C\) and \(A B C^{\prime}\) are reflections of each other in \(A B\) and so are congruent.

Note: The three angles of any triangle are determined

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-09(3)}
by the lengths of the three sides.

\section*{The AAS congruence test}
We first consider two important ideas about triangles.

\section*{Sides opposite angles}
Here is a triangle with the angles marked.

We say that the side \(A B\) is opposite the angle \(40^{\circ}\) at \(C\).

The side \(B C\) is opposite the angle \(30^{\circ}\) at \(A\).

The side \(A C\) is opposite the angle \(110^{\circ}\) at \(B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-09(1)}
\end{center}

\section*{Matching sides and angles}
Here are two triangles with the same three angles in both. We see that \(A\) matches \(P, B\) matches \(Q\) and \(C\) matches \(R\).

We also say that \(A B\) and \(P Q\) are matching sides because:
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-09(2)}

\begin{itemize}
  \item \(A B\) is opposite the angle \(45^{\circ}\) in \(\triangle A B C\), and
\end{itemize}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-09}
\end{center}

\begin{itemize}
  \item \(P Q\) is opposite the angle \(45^{\circ}\) in \(\triangle P Q R\).
\end{itemize}

Similarly, \(B C\) and \(Q R\) are matching sides because they are both opposite \(75^{\circ}\).

Also \(C A\) and \(R P\) are matching sides because they are both opposite \(60^{\circ}\).

\section*{Matching sides}
When two triangles each have the same three angles, we say that a side of one triangle matches a side in the second if the two sides are opposite the same angle.

Notice that if two angles of one triangle are equal to two angles of another, then the third pair of angles are also equal. This is because the angle sum of a triangle is \(180^{\circ}\). For example, the third angle in each triangle below is \(90^{\circ}\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-10(3)}

Thus \(B C\) and \(P Q\) are matching sides in this situation also. (They are matching because they are opposite equal angles.)

Now consider the diagrams below.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-10(5)}

The two triangles above both have angles of \(30^{\circ}, 50^{\circ}\) and \(100^{\circ}\), but they are not congruent, because they have different sizes. This shows that just knowing that three pairs of angles are equal is not enough to establish congruence.

If, however, we know also that a pair of matching sides are equal, then the two triangles are congruent. This test for congruence is called the AAS congruence test. The initials AAS stand for Angle, Angle, Side.

\section*{AAS congruence test}
If two angles and one side of one triangle are equal to two angles and the matching side of another triangle, then the two triangles are congruent.

\section*{Example 2}
In each part, write down a statement that the two triangles are congruent, giving the appropriate congruence test as a reason.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-10(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-10(1)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-10}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-10(4)}
\end{center}

\section*{Solution}
a In the triangles \(A B C\) and \(P Q R\) :

\[
\begin{aligned}
\angle A & =\angle R=30^{\circ} \\
\angle C & =\angle P=70^{\circ} \\
A C & =P R(\text { matching side }) \\
\text { so } \triangle A B C & \equiv \triangle R Q P(\mathrm{AAS})
\end{aligned}
\]

(Note that the second triangle is named in matching order.

\(A\) matches \(R, B\) matches \(Q\) and \(C\) matches \(P\).)

b In the triangles \(A B C\) and \(P Q R\) :

\[
\begin{aligned}
\angle B & =\angle Q=110^{\circ} \text { (given) } \\
\angle C & \left.=40^{\circ} \text { (angle sum of } \triangle A B C\right) \\
\text { so } \angle C & =\angle P \\
B C & =P Q(\text { given }) \\
\text { so } \triangle A B C & \equiv \triangle R Q P(\mathrm{AAS})
\end{aligned}
\]

To see why the AAS test is valid, we look at a particular case of a triangle \(A B C\) with base \(B C=13 \mathrm{~cm}\) and base angles equal to \(B=35^{\circ}\) and \(C=45^{\circ}\). How many such triangles are there? Are they all congruent to one another? We can construct two such triangles by drawing the rays shown and extending them to meet at the points \(A\) and \(A^{\prime}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-11}
\end{center}

Note: \(\triangle A^{\prime} B C\) is the reflection of \(\triangle A B C\) in the line \(B C\) because reflection preserves angles. So these two triangles are congruent.

\section*{Using congruence to find lengths and angles}
Once the congruence of two triangles has been established, we can draw conclusions about the remaining pairs of matching angles or matching sides. This allows us to draw conclusions about lengths and angles.

\section*{Example 3}
Use congruence to find the value of \(\theta\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-12}

\section*{Solution}
From the diagram, \(\triangle A B C \equiv \triangle P Q R(\mathrm{SSS})\)

so \(\theta=\angle B\) (matching angles of congruent triangles)

\(=47^{\circ}\)

\section*{Example 4}
Use congruence to find the value of \(x\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-12(1)}

\section*{Solution}
From the diagram, \(\Delta F G H \equiv \Delta T S R\) (AAS)

so \(x=H G\) (matching sides of congruent triangles)

\(=29 \mathrm{~cm}\)

You may have noticed that the left-hand diagrams in Examples 3 and 4 have too much information. In both triangles, at least one measurement must be only approximate. All of this can safely be ignored at this stage.

\section*{Using congruence to show that two lengths or two angles are equal}
Congruence can be used to show that two matching lengths or two matching angles are equal.

\section*{Example 5}
Use congruence to show that \(\angle P A Q=\angle P B Q\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-13}
\end{center}

\section*{Solution}
From the diagram, \(\triangle P A Q \equiv \triangle P B Q(\mathrm{SSS})\)

so \(\triangle P A Q=\triangle P B Q\) (matching angles of congruent triangles)

Note: The side \(P Q\) is called common to both triangles, and is marked on the diagram with the wavy symbol . In the above example it provides the third pair of equal sides.

\section*{Exercise 12C}
Question 1 is for class discussion.

1 a Three metal rods, of lengths \(2 \mathrm{~m}, 3 \mathrm{~m}\) and \(4 \mathrm{~m}\), are bolted together with hinges at the corners. What congruence test explains why the structure is rigid?

b A rope is attached to the top of a vertical pole. The rope is then tied down to the ground so that it makes an angle of \(20^{\circ}\) to the vertical. Another identical pole nearby is secured in the same way. Do the two ropes necessarily have the same length? Why or why not?

2 a Using a ruler and compasses only, construct a triangle with side lengths of \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(5 \mathrm{~cm}\).

b i What happens when you try to construct a triangle with side lengths of \(4 \mathrm{~cm}, 5 \mathrm{~cm}\) and \(12 \mathrm{~cm}\) ?

ii Copy and complete: 'The longest side of a triangle ...'

c Jack says that in the Great Outback there are three towns \(A, B\) and \(C\). Town \(B\) is \(34 \mathrm{~km}\) from \(A\) and \(68 \mathrm{~km}\) from \(C\). Town \(A\) is \(110 \mathrm{~km}\) from \(C\). (All distances are as the crow flies.) Comment on Jack's claim.

3 a Using a ruler and compasses only, construct a triangle \(A B C\) in which \(A B=12 \mathrm{~cm}\), \(\angle A=45^{\circ}\) and \(\angle B=60^{\circ}\). Measure the length of \(A C\). Which congruence test tells you that your neighbour gets the same length?

b Ajun claims to have constructed a triangle in which one side is \(10 \mathrm{~cm}\) and the angles at each end of this side are \(105^{\circ}\) and \(80^{\circ}\). Comment on Ajun's claim.

4 In each part below, say whether the two triangles are congruent. If they are congruent, write a congruence statement, including the appropriate congruence test.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(17)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(6)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(15)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(3)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(12)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(10)}
\end{center}

d \({ }_{T}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(7)}
\end{center}

f \(Q\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(5)}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-14(8)}
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-14(11)}

5 If possible, write down a congruence statement (with the appropriate congruence test) in each part. Then find the value of \(x\) or \(\theta\), giving reasons. Keep the vertices written in matching order.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(4)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(13)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(14)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(16)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-14(2)}
\end{center}

6 In each part, first write a congruence statement. Then prove the required result, giving all reasons.

a Prove that \(\angle A C B=\angle V P T\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-14(9)}

b Prove that \(F H=E M\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-14(1)}

\section*{12 Congruent triangles: The SAS and RHS tests}
We saw in the preceding section that having two pairs of equal sides is not enough to establish that two triangles are congruent, because the two sides can flap about freely, like the blades of a pair of scissors.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-15(1)}

The two tests discussed in this section stop the two sides flapping about by specifying one angle of the triangle as well.

\section*{The SAS congruence test}
The easiest way to stop the two sides flapping is to specify the angle between them. This angle is called the included angle. ('Included' means 'shut in'.) The resulting test is called the SAS congruence test. The letters SAS stand for Side, Angle, Side.

\section*{SAS congruence test}
If two sides of one triangle are respectively equal to two sides of another triangle, and the angles included between the sides in each pair are equal, then the triangles are congruent.

Be careful! For the test to work, the angles must be the ones included between the sides in each pair.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-15}

In the diagram above, \(\triangle A B C \equiv \triangle P Q R(\mathrm{SAS})\).

To see why this is true, take a triangle \(A B C\) with side lengths \(A B=4 \mathrm{~cm}, B C=6 \mathrm{~cm}\) and included angle, \(\angle B=30^{\circ}\). Its reflection in the line \(B C\) is also shown. Triangle \(A B C\) is congruent to triangle \(A^{\prime} B C\) because reflection in \(B C\) moves \(A\) to \(A^{\prime}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-15(2)}
\end{center}

Check for yourself that any other triangle in the plane with the given side lengths and included angle is one of the triangles we have drawn with \(B C\) in a different position and orientation.

\section*{Example 6}
Write down a statement that the two triangles below are congruent, giving the appropriate congruence test as a reason.
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-16(1)}

\section*{Solution}
\(\Delta H F G \equiv \triangle R T S(\mathrm{SAS})\)

\section*{The RHS congruence test}
A useful test that can be applied to right-angled triangles, and is easy to apply, is the

RHS congruence test. (The letters RHS stand for Right angle, Hypotenuse, Side.)

\section*{RHS congruence test}
If two right-angled triangles have equal hypotenuses, and another pair of equal sides, then the triangles are congruent.

\section*{Proof of the RHS congruence test}
\[
\begin{aligned}
B C^{2} & =A C^{2}-A B^{2}(\text { Pythagoras' theorem in } \triangle A B C) \\
& =P Q^{2}-P R^{2} \text { (equal hypotenuse and equal side) } \\
& \left.=Q R^{2} \text { (right-angled } \triangle P Q R\right)
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-16}
\end{center}

So \(B C=Q R\) and the two triangles have equal sides. Hence, they are congruent by the SSS congruence test.

\section*{Constructing a right-angled triangle with a given hypotenuse and side}
Construct a right-angled triangle \(A B C\) with \(\angle A=90^{\circ}, B C=13 \mathrm{~cm}\) and \(A B=12 \mathrm{~cm}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-17}
\end{center}

The side \(B C\) is the hypotenuse of the right-angled triangle. By Pythagoras' theorem, it must be the longest side.

Step 1: In the middle of your page, draw an interval \(A B\) of length \(12 \mathrm{~cm}\).

Step 2: Draw a line through \(A\) at right angles to \(A B\).

Step 3: With centre \(B\) and radius \(13 \mathrm{~cm}\), draw a circle crossing the vertical line at two points \(C\) and \(C^{\prime}\).

Step 4: Join up the triangle \(A B C\) and the triangle \(A B C^{\prime}\).

The two triangles you have drawn are congruent, being reflections of each other in \(A B\).

\section*{Example 7}
Write down a statement that the two triangles below are congruent, giving the appropriate congruence test as a reason.

\section*{Solution}
\(\triangle A O V \equiv \triangle F O V(\mathrm{RHS})\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-17(1)}
\end{center}

\section*{The four standard congruence tests for triangles}
Two triangles are congruent if:

SSS: the three sides of one triangle are respectively equal to the three sides of the other triangle, or

AAS: two angles and one side of one triangle are respectively equal to two angles and the matching side of the other triangle, or

SAS: two sides and the included angle of one triangle are respectively equal to two sides and the included angle of the other triangle, or

RHS: the hypotenuse and one side of one right-angled triangle are respectively equal to the hypotenuse and one side of the other right-angled triangle.

\section*{The case of the non-included angle}
In the diagram, triangles \(B C A\) and \(B C D\) have a common side \(B C\), a common angle \(\angle B\), and \(A C=D C\). As you can see, the triangles are not congruent.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-18}
\end{center}

Therefore two sides and a non-included angle are not sufficient to determine congruency. Further discussion of the case of the non-included angle is to be found in Question 6 of Exercise 12D and Question 8 of the Challenge exercise.

\section*{Using congruence to find sides or angles}
When two triangles are congruent, and we are given side lengths and/or angles of one of them, we can find the matching sides and/or angles of the other.

\section*{Example 8}
Use congruence to prove \(\theta=\alpha\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-18(1)}

\(F\)

\section*{Solution}
From the diagram, \(\triangle F G H \equiv \triangle T S R(\mathrm{RHS})\)

so \(\theta=\alpha\) (matching angles of congruent triangles)

\section*{Example 9}
Use congruence to show that \(G H=L M\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-19(2)}

\section*{Solution}
From the diagram, \(\triangle G F H \equiv \triangle L K M\) (SAS)

so \(G H=L M\) (matching sides of congruent triangles)

\section*{Exercise 12D}
Question 1 is suitable for discussion.

1 a Two pairs of compasses, each with legs \(12 \mathrm{~cm}\) long, are opened so that the arms make an angle of \(25^{\circ}\). What congruence test explains why the distance between the point and the pencil lead is the same in both compasses?

b Two 6-metre ladders on horizontal ground rest against a wall, with each ladder reaching 5 metres up the wall. What congruence test explains why the angle between the ladder and the ground is the same for both ladders?

c Pradap attached a rope 9 metres up a vertical pole, then stretched the rope and secured it to the ground near the pole. He measured the length of the rope to be 8.5 metres.

Comment on this situation.

2 a Using a ruler and compasses only, construct an isosceles triangle with legs that are \(8 \mathrm{~cm}\) each and with an apex angle that is \(45^{\circ}\). Measure the length of the base.

b Using ruler and compasses only, construct a right-angled triangle \(A B C\) in which \(\angle A\) is a right angle, \(A B=6 \mathrm{~cm}\) and the hypotenuse \(B C=8 \mathrm{~cm}\). Use your protractor to measure the size of \(\angle B\).

3 In each part below, if the two triangles are congruent, write down the congruence statement, giving the appropriate congruence test as a reason.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-19}
\end{center}

b \(R\)

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-19(1)}
\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20(9)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20(7)}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-20(5)}

f \(\mathrm{C}\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-20(2)}

4 If possible, write down a congruence statement (with the appropriate congruence test). Then find the value of \(x\) or \(\theta\), giving the reason.

a
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-20(4)}

\(\mathbf{c}\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-20(8)}

b \(c\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20(1)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20}
\end{center}

5 In each part, first write a congruence statement. Then prove the required result, giving all reasons.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20(6)}
\end{center}

Prove that \(\angle M=\angle T\).

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20(10)}
\end{center}

Prove that \(I J=K L\). b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20(3)}
\end{center}

Prove that \(W L=M L\).

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-20(11)}
\end{center}

Prove that \(H T=G W\).

6 The SAS congruence test requires that the angle be the angle included between the two sides. Here is an example where the angle is not included - the construction will produce two triangles that are not congruent. Construct a triangle \(A B C\) with \(A C=8 \mathrm{~cm}, B C=6 \mathrm{~cm}\) and \(\angle A=30^{\circ}\) as follows.

Step 1: Draw a long horizontal line \(A X\) near the bottom of a new page in your exercise book.

Step 2: Construct an angle of \(30^{\circ}\) at \(A\).

Step 3: Mark \(C\) on the sloping arm so that \(A C=8 \mathrm{~cm}\).

Step 4: With centre \(C\) and radius \(6 \mathrm{~cm}\), draw an arc cutting the horizontal line at \(B\) and \(B^{\prime}\).

Step 5: Join \(C B\) and \(C B^{\prime}\).

Then \(\triangle A B C\) and \(\triangle A B^{\prime} C\) both fulfil the conditions, but they are clearly not congruent.

\section*{12 Using congruence in geometrical problems}
This section shows how to use congruence in geometrical reasoning and problems.

\section*{Setting out congruence proofs}
Here are three examples showing a recommended way of writing out a solution to a problem involving congruence. Attention to detail is critical.

\section*{Example 10}
Find the length of \(A P\) in the diagram below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-21}
\end{center}

\section*{Solution}
In the triangles \(A B P\) and \(B C Q\) :

\[
\begin{aligned}
A B & =B C(\text { given }) \\
\angle B A P & =\angle C B Q(\text { corresponding angles, } A P \| B Q) \\
\angle A B P & =\angle B C Q(\text { corresponding angles, } B P \| C Q) \\
\text { so } \triangle A B P & \equiv \triangle B C Q(\text { AAS })
\end{aligned}
\]

Hence, \(A P=12 \mathrm{~cm}\) (matching sides of congruent triangles)

\section*{Proving that two lines are parallel}
Congruence is often used to prove that two alternating angles or two corresponding angles are equal. It then follows that the associated lines are parallel.

\section*{Example 11}
Prove that \(A B \| P Q\) in the diagram below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-22}
\end{center}

\section*{Solution}
In the triangles \(A O B\) and \(Q O P\) :

\[
\begin{aligned}
A O & =Q O(\text { given }) \\
B O & =P O(\text { given }) \\
\angle A O B & =\angle Q O P(\text { vertically opposite at } O) \\
\text { so } \triangle A O B & \equiv \triangle Q O P(\text { SAS }) \\
\text { Hence } \angle A & =\angle Q \text { (matching angles of congruent triangles) } \\
\text { so } A B \| P Q & \text { (alternate angles are equal) }
\end{aligned}
\]

\section*{Proving that two lines are perpendicular}
If two angles are equal and add to \(180^{\circ}\), they must both be right angles. This idea is often used in problems to prove that two lines are perpendicular.

\section*{Example 12}
In the diagram opposite, \(F M\) joins the apex \(F\) of the isosceles \(\triangle F G H\) to the midpoint \(M\) of its base. Use congruence to prove that \(F M \perp G H\); that is, \(F M\) is perpendicular to \(G H\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-23(1)}
\end{center}

\section*{Solution}
In the triangles GFM and \(H F M\) :

\[
\begin{aligned}
F M & =F M(\text { common }) \\
F G & =F H(\text { given }) \\
G M & =H M(\text { given }) \\
\text { so } \Delta G F M & =\Delta H F M(\mathrm{SSS})
\end{aligned}
\]

Hence \(\angle G M F=\angle H M F\) (matching angles of congruent triangles)

But \(\angle G M F+\angle H M F=180^{\circ}(\) straight angle at \(M)\)

\[
\text { so } \begin{aligned}
\angle G M F & =\angle H M F \\
& =90^{\circ}
\end{aligned}
\]

\section*{Exercise 12E}
1 In each part below, prove that two triangles are congruent. Hence, find the value of \(x\) or \(\theta\). The point \(O\) is always the centre of the circle.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-23(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-23}
\end{center}

c

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-23(3)}
d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(4)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(6)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(2)}
\end{center}

2 In these problems, the points \(O\) and \(Z\) are always the centres of circles.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(1)}
\end{center}

i Prove that \(\triangle R L S \equiv \triangle R M S\).

ii Hence, show that \(L S=M S\).

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(7)}
\end{center}

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

\(\mathbf{e}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(3)}
\end{center}

Prove that \(\angle O F H=\angle O I G\). b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(5)}
\end{center}

i Prove that \(\triangle C O Z \equiv \triangle D O Z\).

ii Hence, show that \(\angle O C Z=\angle O D Z\).

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24}
\end{center}

Prove that \(S N=T N\).

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-24(8)}
\end{center}

Prove that \(V W=F G\).

Example 11 3 In each part below, use congruent triangles to prove that the lines \(A B\) and \(P Q\) are parallel. The point \(O\) is always the centre of the circle.

\(\mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25(7)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25(6)}
\end{center}

4 In each part below, use congruent triangles to prove that the lines \(A B\) and \(P Q\) are perpendicular. The point \(O\) is always the centre of the circle.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25(3)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25(5)}
\end{center}

5 Each of these problems requires two applications of congruence. The points \(O\) and \(Z\) are the centres of circles.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25(1)}
\end{center}

i Prove that \(\triangle G D M \equiv \triangle E D M\).

ii Hence, show that \(D M \perp G E\).

iii Prove that \(\triangle G M F \equiv \triangle E M F\).

iv Hence, show that \(G F=E F\). \(D E F G\) is called a kite. b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-25(4)}
\end{center}

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

ii Hence, show that \(\angle A O W=\angle B O W\).

iii Prove that \(\triangle A O W \equiv \triangle B O W\).

iv Hence, show that \(O Z\) is the perpendicular bisector of \(A B\).

\section*{12 Congruence and special triangles}
Isosceles and equilateral triangles were introduced in Chapter 13 of ICE-EM Mathematics Year 7. Congruence allows us to give proper proofs of three basic theorems about these special triangles.

First, here is a reminder of the definitions of isosceles and equilateral triangles.

\section*{Isosceles and equilateral triangles}
\begin{itemize}
  \item An isosceles triangle is a triangle with two (or more) sides equal.
  \item The equal sides are called the legs and the third side is called the base.
  \item The legs meet at the apex and the other two angles are the base angles.
  \item An equilateral triangle is a triangle with all three sides equal.
\end{itemize}

The word 'isosceles' comes from Greek and means 'equal legs'. The word 'equilateral' comes from Latin and means 'equal sides'.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-26(1)}
\end{center}

Isosceles triangle

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-26}
\end{center}

Equilateral triangle

\section*{The base angles of an isosceles triangle are equal}
Notice that the proof is based on congruent triangles.

Theorem: The base angles of an isosceles triangle are equal.

Proof: \(\quad\) Let \(\triangle A B C\) be isosceles, with \(C A=C B\).

We need to prove that \(\angle \mathrm{A}=\angle B\).

Draw the bisector of \(\angle A C B\), and let it meet \(A B\) at \(M\).

In the triangles \(A C M\) and \(B C M\) :

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-27}
\end{center}

\[
\begin{aligned}
C A & =C B(\text { given }) \\
C M & =C M(\text { common side }) \\
\angle A C M & =\angle B C M(\text { construction }) \\
\text { so } \triangle A C M & \equiv \triangle B C M(\text { SAS })
\end{aligned}
\]

Hence \(\angle \mathrm{A}=\angle B\) (matching angles of congruent triangles)

\section*{If two angles of a triangle are equal, then the sides opposite those angles are equal}
This proof is also based on congruence.

Theorem: If two angles of a triangle are equal, then the sides opposite those angles are equal. (That is, the triangle is isosceles.)

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

We need to prove that \(C A=C B\).

Draw the bisector of \(\angle A C B\), and let it meet \(A B\) at \(M\).

In the triangles \(A M C\) and \(B M C\) :

\[
\begin{aligned}
\angle C A M & =\angle C B M(\text { given }) \\
\angle A C M & =\angle B C M(\text { construction }) \\
C M & =C M(\text { common side })
\end{aligned}
\]

so \(\triangle A M C \equiv \triangle B M C\) (AAS)

Hence \(C A=C B\) (matching sides of congruent triangles)

That is, the triangle is isosceles.

This second theorem is the converse of the first theorem.

\begin{itemize}
  \item The first theorem says: 'If two sides of a triangle are equal, then the angles opposite those sides are equal'.
  \item The second theorem says: 'If two angles of a triangle are equal, then the sides opposite those angles are equal'.
\end{itemize}

We have now proved that both the theorem and its converse are true.

\section*{Isosceles triangles}
\begin{itemize}
  \item The base angles of an isosceles triangle are equal.
  \item Conversely, if two angles of a triangle are equal, then the sides opposite those angles are equal. (That is, the triangle is isosceles.)
\end{itemize}

\section*{The angles of an equilateral triangle are all \(60^{\circ}\)}
We can prove this result, by applying the earlier result that the base angles of an isosceles triangle are equal. This result is a simple corollary of the earlier result that the base angles of an isosceles triangle are equal.

Theorem: The interior angles of an equilateral triangle are all \(60^{\circ}\).

Proof:

\[
\begin{array}{clrl}
A C & =A B(\text { given }) \\
\text { so } & \angle C & =\angle B(\text { angles opposite equal sides }) \\
\text { Also } & C A & =C B(\text { given }) \\
\text { so } & \angle A & =\angle B(\text { angles opposite equal sides })
\end{array}
\]

Therefore \(\angle A=\angle B=\angle C\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-28}
\end{center}

Since the angles add to \(180^{\circ}\), each angle must be \(180 \div 3=60^{\circ}\).

\section*{Equilateral triangles}
\begin{itemize}
  \item Each interior angle of an equilateral triangle is \(60^{\circ}\).
  \item Conversely, if all the angles of a triangle are equal, then the triangle is equilateral.
\end{itemize}

The last dot point is proven in Question 2 of Exercise 12F.

\section*{Example 13}
a Find \(\alpha\) and \(\beta\) in the diagram below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-28(2)}
\end{center}

b Find \(x\) and \(\angle P S Q\) in the diagram below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-28(1)}
\end{center}

\section*{Solution}
a First, \(\quad \alpha=60^{\circ}\) (equilateral \(\triangle A B C\) )

Also, \(\angle D B A=\beta\) (base angles of isosceles \(\triangle A B D\) )

so \(\beta+\beta+40^{\circ}=180^{\circ}(\) angle sum of \(\triangle D A B)\)

Hence \(\quad \beta=70^{\circ}\)

b First, \(\angle P S R=60^{\circ}\) (equilateral \(\triangle A B C\) )

Also, \(\angle P S Q=94^{\circ}\) (adjacent angles)

so \(\quad R Q=S Q\) (opposite sides of \(\triangle Q R S\) are equal),

Hence \(\quad x=6 \mathrm{~cm}\)

\section*{Exercise 12F}
1 Use the results of this section about isosceles and equilateral triangles to find the values of \(x, y, \alpha, \beta, \gamma\) and \(\theta\). Give all reasons in your solutions. The point \(O\) in part \(\mathbf{n}\) is the centre of the circle.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(6)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(5)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(11)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(8)}
\end{center}

\(\mathbf{k}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(1)}
\end{center}

b \(R\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(3)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(9)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(7)}
\end{center}

\(\mathbf{f}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(4)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(10)}
\end{center}

j

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-29(2)}
\end{center}

I

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-29}
m

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-30(4)}
\end{center}

n

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-30(2)}
\end{center}

2 In \(\triangle A B C\) opposite, all angles are equal.

a Explain why all the angles are \(60^{\circ}\).

b Explain why the triangle is equilateral.

You have now proven a test for an equilateral triangle:

'If all angles of a triangle are equal, then the triangle is equilateral'.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-30}
\end{center}

3 Let \(\triangle A B C\) be isosceles, with \(A C=B C\), and join \(C M\), where \(M\) is the midpoint of the base \(A B\).

a Show that \(\triangle A M C \equiv \triangle B M C\).

b Hence, show that \(\angle A=\angle B\).

This is a second proof of 'the base angles of an isosceles triangle are equal'.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-30(6)}
\end{center}

4 Let \(A B C\) be a triangle with \(\angle A=\angle B\). We will prove that \(A C=B C\).

Construct the line through \(C\) perpendicular to the base \(A B\).

Let this line meet \(A B\) at \(N\).

a Show that \(\triangle A N C \equiv \triangle B N C\).

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

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-30(5)}
\end{center}

This is another proof of the result 'If the two base angles of a triangle are equal then the sides opposite those angles are equal'.

\(5 \triangle A B C\) is an isosceles triangle with \(A B=A C\).

The angle bisector of \(\angle B\) intersects the angle bisector of \(\angle C\) at \(X\).

Prove that \(\angle B X C=\angle A C D\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-30(1)}
\end{center}

\(6 \triangle A B C\) is an isosceles triangle with \(A B=A C\). \(X\) is the point on \(A C\) such that \(B X\) is perpendicular to \(A C\).

Prove that \(\angle X B C=\frac{1}{2} \angle B A C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-30(3)}
\end{center}

7 The bisector of an angle of a triangle cuts the opposite side at right angles. Prove that the triangle is isosceles.

\section*{Review exercise}
1 In each part write a congruence statement and give a reason.

a \(C\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-31(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-31(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-31(1)}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-31(2)}
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-31(7)}

2 a In the diagram opposite, name the two congruent triangles and explain why they are congruent.

b Find \(A C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-31(9)}
\end{center}

3 For each pair, write a congruence statement and give a reason.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-31}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-31(4)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-31(8)}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-31(5)}
c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(4)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(6)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(9)}
\end{center}

4 In each part, name the triangle congruent to \(\triangle A B C\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(5)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(10)}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-32(1)}
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-32(11)}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(7)}
\end{center}

5 In each part, name all the pairs of congruent triangles, giving the abbreviated reason.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-32(8)}
\end{center}

6 Find the values of the pronumerals in each.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-33(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-33(5)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-33(2)}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-33(3)}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-33}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-33(1)}
\end{center}

\(b \mathrm{~cm}\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-33(7)}

\section*{Challenge exercise}
1 This question provides a proof of the angle bisector construction. The arcs intersecting at \(M\) are centred at \(P\) and \(Q\).

a Copy the diagram and join the intervals \(P M\) and \(Q M\).

b Prove that \(\triangle P M O \equiv \triangle Q M O\).

c Hence explain why \(O M\) is the bisector of \(\angle A O B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-33(6)}
\end{center}

2 This question provides a proof of the construction of the perpendicular bisector of an interval. The arcs are of the same radius and centred at \(A\) and \(B\).

a Copy the diagram and join the intervals \(A P, B P, A Q\) and \(B Q\).

b Prove that \(\triangle A P Q \equiv \triangle B P Q\).

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

d Prove that \(\triangle A P M \equiv \triangle B P M\).

e Hence prove that \(A M=B M\) and \(P Q \perp A B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-34}
\end{center}

3 This question provides a proof of the construction of the perpendicular from a given point to a given line.

a Copy the diagram and join the intervals \(A F, B F, A G\), \(B G\) and \(F G\).

b Prove that \(\triangle A F G \equiv \triangle B F G\).

c Hence prove that \(\angle A F G=\angle B F G\).

d Prove that \(\triangle A F M \equiv \triangle B F M\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-34(2)}
\end{center}

e Hence prove that \(F G \perp A B\).

4 In this question, we prove that the base angles of an isosceles triangle are equal by proving that the triangle is congruent to itself. The proof is reputedly due to the Greek mathematician Pappus, who lived in Alexandria from about \(290 \mathrm{CE}\) to about \(350 \mathrm{CE}\). Let \(\triangle A B C\) be isosceles, with \(A C=B C\).

a Prove that \(\triangle A C B \equiv \triangle B C A\). (Note the changed order of the vertices.)

b Hence show that \(\angle A=\angle B\).

5 In this question, we give another proof that a triangle with equal base angles is isosceles. This proof involves no construction - the triangle is proven to be congruent to itself. Let \(A B C\) be a triangle with \(\angle A=\angle B\).

We just prove that \(A C=B C\).

a Prove that \(\triangle A B C \equiv \triangle B A C\).

b Hence show that \(A C=B C\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b04970f64b9f160d323fg-34(1)}

6 Explain why the supposed 'proof' outlined below is invalid.

Let \(A B C\) be a triangle with \(\angle A=\angle B\).

Mark the midpoint \(M\) of the base \(A B\), and join \(C M\).

Then \(\triangle A M C \equiv \triangle B M C\).

Hence \(A C=B C\).

7 The result 'proven' below is clearly false.

Explain why the reasoning is invalid.

Let \(\triangle A B C\) be an isosceles triangle with \(A C=B C\).

Let \(P\) be any point on the base \(A B\), and join \(P C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-35}
\end{center}

In the triangles \(A C P\) and \(B C P\) :

\[
\begin{aligned}
C P & =C P & & \text { (common) } \\
C A & =C B & & \text { (given) } \\
\angle C A P & =\angle C B P & & \text { (base angles of isosceles } \triangle A B C) \\
\text { so } \triangle A C P & \equiv \triangle B C P & & \text { (SAS) } \\
\text { Hence } A P & =B P & & \text { (matching sides of congruent triangles) }
\end{aligned}
\]

so \(P\) is the midpoint of \(A B\).

8 There is a valid 'OSS congruence test' (where O stands for 'obtuse angle'):

'If two sides and an obtuse non-included angle of one triangle are equal to two sides and a matching obtuse non-included angle of another triangle, then the triangles are congruent.'

Explain why this result is true.

9 The Greeks said that the oldest geometrical theorem stated and proved was Thales' theorem. Thales' theorem states:

'An angle in a semicircle is a right angle.'

a Let \(A O B\) be a diameter of a circle with centre \(O\). Let \(P\) be any other point on the circumference. Join \(A P, B P\) and \(O P\), and let \(\angle P A B=\alpha\) and \(\angle P B A=\beta\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b04970f64b9f160d323fg-35(1)}
\end{center}

b Use isosceles triangles to prove that \(\angle A P B=90^{\circ}\).

10 The lines through each vertex of a triangle perpendicular to the opposite side are called the altitudes of a triangle. They are concurrent at the orthocentre of the triangle.

(This can be proved using theorems that you will meet in Year 10.) The orthocentre may lie outside the triangle.

a On a large sheet of paper, draw an obtuse-angled triangle, then construct its orthocentre. You will need to produce two of the sides to perform this construction.

b Let \(H\) be the orthocentre of \(\triangle A B C\). Prove that each of the four points \(A, B, C\) and \(H\) is the orthocentre of the triangle formed by the other three points. This is true in both cases - when \(\triangle A B C\) is acute-angled and when \(\triangle A B C\) is obtuse-angled.

11 The perpendicular bisectors of the sides of a triangle are concurrent at a point called the circumcentre, and this point is the centre of the circle through all three vertices. Here is an outline of a proof. Check the details.

a Take a triangle \(A B C\). Let \(P\) be the midpoint of \(B C, Q\) be the midpoint of \(C A\), and \(R\) be the midpoint of \(A B\). Let the perpendicular bisectors of \(A B\) and \(A C\) meet at \(M\). Join \(M A, M B\) and \(M C\).

b Prove that \(\triangle A M R \equiv \triangle B M R\) and that \(\triangle A M Q \equiv \triangle C M Q\).

c Hence prove that the circle with centre \(M\) and radius \(M A\) passes through \(B\) and \(C\).

d Prove that \(M P\) is perpendicular to \(B C\).

e Under what circumstances does the circumcentre lie outside the triangle?


\end{document}