## Chapter 07 Triangles

### 7.1 Introduction

You have studied about triangles and their various properties in your earlier classes. You know that a closed figure formed by three intersecting lines is called a triangle. (‘Tri’ means ’three’). A triangle has three sides, three angles and three vertices. For example, in triangle $\mathrm{ABC}$, denoted as $\triangle \mathrm{ABC}$ (see Fig. 7.1); $\mathrm{AB}, \mathrm{BC}, \mathrm{CA}$ are the three sides, $\angle \mathrm{A}, \angle \mathrm{B}, \angle \mathrm{C}$ are the three angles and $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are three vertices. In Chapter 6, you have also studied some properties of triangles. In this chapter, you will study in details about the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle. You have already verified most of these properties in earlier classes. We will now prove some of them.

### 7.2 Congruence of Triangles

Fig. 7.1

You must have observed that two copies of your photographs of the same size are identical. Similarly, two bangles of the same size, two ATM cards issued by the same bank are identical. You may recall that on placing a one rupee coin on another minted in the same year, they cover each other completely.

Do you remember what such figures are called? Indeed they are called congruent figures (‘congruent’ means equal in all respects or figures whose shapes and sizes are both the same).

Now, draw two circles of the same radius and place one on the other. What do you observe? They cover each other completely and we call them as congruent circles.

Repeat this activity by placing one square on the other with sides of the same measure (see Fig. 7.2) or by placing two equilateral triangles of equal sides on each other. You will observe that the squares are congruent to each other and so are the equilateral triangles.

Fig. 7.2

You may wonder why we are studying congruence. You all must have seen the ice tray in your refrigerator. Observe that the moulds for making ice are all congruent. The cast used for moulding in the tray also has congruent depressions (may be all are rectangular or all circular or all triangular). So, whenever identical objects have to be produced, the concept of congruence is used in making the cast.

Sometimes, you may find it difficult to replace the refill in your pen by a new one and this is so when the new refill is not of the same size as the one you want to remove. Obviously, if the two refills are identical or congruent, the new refill fits.

So, you can find numerous examples where congruence of objects is applied in daily life situations.

Can you think of some more examples of congruent figures?

Now, which of the following figures are not congruent to the square in Fig 7.3 (i) :

Fig. 7.3

The large squares in Fig. 7.3 (ii) and (iii) are obviously not congruent to the one in Fig 7.3 (i), but the square in Fig 7.3 (iv) is congruent to the one given in Fig 7.3 (i).

Let us now discuss the congruence of two triangles.

You already know that two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.

Now, which of the triangles given below are congruent to triangle $\mathrm{ABC}$ in Fig. 7.4 (i)?

Fig. 7.4

Cut out each of these triangles from Fig. 7.4 (ii) to (v) and turn them around and try to cover $\triangle \mathrm{ABC}$. Observe that triangles in Fig. 7.4 (ii), (iii) and (iv) are congruent to $\triangle \mathrm{ABC}$ while $\triangle \mathrm{TSU}$ of Fig 7.4 (v) is not congruent to $\triangle \mathrm{ABC}$.

If $\triangle \mathrm{PQR}$ is congruent to $\triangle \mathrm{ABC}$, we write $\Delta \mathrm{PQR} \cong \triangle \mathrm{ABC}$.

Notice that when $\Delta \mathrm{PQR} \cong \triangle \mathrm{ABC}$, then sides of $\Delta \mathrm{PQR}$ fall on corresponding equal sides of $\triangle \mathrm{ABC}$ and so is the case for the angles.

That is, $\mathrm{PQ}$ covers $\mathrm{AB}, \mathrm{QR}$ covers $\mathrm{BC}$ and $\mathrm{RP}$ covers $\mathrm{CA} ; \angle \mathrm{P}$ covers $\angle \mathrm{A}$, $\angle \mathrm{Q}$ covers $\angle \mathrm{B}$ and $\angle \mathrm{R}$ covers $\angle \mathrm{C}$. Also, there is a one-one correspondence between the vertices. That is, $\mathrm{P}$ corresponds to $\mathrm{A}, \mathrm{Q}$ to $\mathrm{B}, \mathrm{R}$ to $\mathrm{C}$ and so on which is written as

$$\mathrm{P} \leftrightarrow \mathrm{A}, \mathrm{Q} \leftrightarrow \mathrm{B},\mathrm{R} \leftrightarrow \mathrm{C}$$

Note that under this correspondence, $\triangle \mathrm{PQR} \cong \triangle \mathrm{ABC}$; but it will not be correct to write $\Delta \mathrm{QRP} \cong \triangle \mathrm{ABC}$.

Similarly, for Fig. 7.4 (iii),

$$\mathrm{FD} \leftrightarrow \mathrm{AB}, \mathrm{DE} \leftrightarrow \mathrm{BC} \text { and } \mathrm{EF} \leftrightarrow \mathrm{CA} $$

$ \text { and } \mathrm{F} \leftrightarrow \mathrm{A}, \mathrm{D} \leftrightarrow \mathrm{B} \text { and } \mathrm{E} \leftrightarrow \mathrm{C}$ So, $\Delta \mathrm{FDE} \cong \triangle \mathrm{ABC}$ but writing $\Delta \mathrm{DEF} \cong \Delta \mathrm{ABC}$ is not correct.

Give the correspondence between the triangle in Fig. 7.4 (iv) and $\triangle \mathrm{ABC}$.

So, it is necessary to write the correspondence of vertices correctly for writing of congruence of triangles in symbolic form.

Note that in congruent triangles corresponding parts are equal and we write in short ’ $\mathrm{CPCT}$ ’ for corresponding parts of congruent triangles.

### 7.3 Criteria for Congruence of Triangles

In earlier classes, you have learnt four criteria for congruence of triangles. Let us recall them.

Draw two triangles with one side $3 \mathrm{~cm}$. Are these triangles congruent? Observe that they are not congruent (see Fig. 7.5).

Fig. 7.5

Now, draw two triangles with one side $4 \mathrm{~cm}$ and one angle $50^{\circ}$ (see Fig. 7.6). Are they congruent?

Fig. 7.6

See that these two triangles are not congruent.

Repeat this activity with some more pairs of triangles.

So, equality of one pair of sides or one pair of sides and one pair of angles is not sufficient to give us congruent triangles.

What would happen if the other pair of arms (sides) of the equal angles are also equal?

In Fig 7.7, $\mathrm{BC}=\mathrm{QR}, \angle \mathrm{B}=\angle \mathrm{Q}$ and also, $\mathrm{AB}=\mathrm{PQ}$. Now, what can you say about congruence of $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$ ?

Recall from your earlier classes that, in this case, the two triangles are congruent. Verify this for $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$ in Fig. 7.7. Repeat this activity with other pairs of triangles. Do you observe that the equality of two sides and the included angle is enough for the congruence of triangles? Yes, it is enough.

Fig. 7.7

This is the first criterion for congruence of triangles.

**Axiom 7.1 (SAS congruence rule) :** Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

This result cannot be proved with the help of previously known results and so it is accepted true as an axiom (see Appendix 1).

Let us now take some examples.

**Example 1 :** In Fig. 7.8, $\mathrm{OA}=\mathrm{OB}$ and $\mathrm{OD}=\mathrm{OC}$. Show that

(i) $\triangle \mathrm{AOD} \cong \triangle \mathrm{BOC}$ and

(ii) $\mathrm{AD} || \mathrm{BC}$.

**Solution :** (i) You may observe that in $\triangle \mathrm{AOD}$ and $\triangle \mathrm{BOC}$,

$OA=OB\quad \text {(Given)}$

$OD=OC\quad \text {(Given)}$

Fig. 7.8

Also, since $\angle \mathrm{AOD}$ and $\angle \mathrm{BOC}$ form a pair of vertically opposite angles, we have

$\quad \quad$ $\angle \mathrm{AOD}=\angle \mathrm{BOC}$.

So, $\quad$ $\triangle \mathrm{AOD} \cong \triangle \mathrm{BOC} \quad$ (by the SAS congruence rule)

(ii) In congruent triangles $\mathrm{AOD}$ and $\mathrm{BOC}$, the other corresponding parts are also equal.

So, $\quad$ $\angle \mathrm{OAD}=\angle \mathrm{OBC}$ and these form a pair of alternate angles for line segments $\mathrm{AD}$ and $\mathrm{BC}$.

Therefore,$\quad$$\mathrm{AD} || \mathrm{BC}$.

**Example 2 :** AB is a line segment and line $l$ is its perpendicular bisector. If a point $\mathrm{P}$ lies on $l$, show that $\mathrm{P}$ is equidistant from $\mathrm{A}$ and $\mathrm{B}$.

**Solution :** Line $l \perp \mathrm{AB}$ and passes through $\mathrm{C}$ which is the mid-point of $A B$ (see Fig. 7.9). You have to show that $\mathrm{PA}=\mathrm{PB}$. Consider $\triangle \mathrm{PCA}$ and $\triangle \mathrm{PCB}$.

We have $\quad \mathrm{AC}=\mathrm{BC} \quad(\mathrm{C}$ is the mid-point of $\mathrm{AB})$

$ \angle \mathrm{PCA}=\angle \mathrm{PCB}=90^{\circ} \text { (Given) } $

$\mathrm{PC} =\mathrm{PC} \quad \text {(Common)}$

So, $\quad \triangle \mathrm{PCA} \cong \triangle \mathrm{PCB}$ (SAS rule)

and so, $\mathrm{PA}=\mathrm{PB}$, as they are corresponding sides of congruent triangles.$^{\circ}$

Fig. 7.9

Now, let us construct two triangles, whose sides are $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ and one of the angles is $50^{\circ}$ and this angle is not included in between the equal sides (see Fig. 7.10). Are the two triangles congruent?

Fig. 7.10

Notice that the two triangles are not congruent.

Repeat this activity with more pairs of triangles. You will observe that for triangles to be congruent, it is very important that the equal angles are included between the pairs of equal sides.

So, SAS congruence rule holds but not ASS or SSA rule.

Next, try to construct the two triangles in which two angles are $60^{\circ}$ and $45^{\circ}$ and the side included between these angles is $4 \mathrm{~cm}$ (see Fig. 7.11).

Fig. 7.11

Cut out these triangles and place one triangle on the other. What do you observe? See that one triangle covers the other completely; that is, the two triangles are congruent. Repeat this activity with more pairs of triangles. You will observe that equality of two angles and the included side is sufficient for congruence of triangles.

This result is the **Angle-Side-Angle** criterion for congruence and is written as ASA criterion. You have verified this criterion in earlier classes, but let us state and prove this result.

Since this result can be proved, it is called a theorem and to prove it, we use the SAS axiom for congruence.

**Theorem 7.1 (ASA congruence rule) :** Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

**Proof :** We are given two triangles $\mathrm{ABC}$ and $\mathrm{DEF}$ in which:

$\angle \mathrm{B}=\angle \mathrm{E}, \angle \mathrm{C}=\angle \mathrm{F}$

and $\quad \mathrm{BC}=\mathrm{EF}$

We need to prove that $\quad \triangle \mathrm{ABC} \cong \triangle \mathrm{DEF}$

For proving the congruence of the two triangles see that three cases arise.

**Case (i) :** Let $\mathrm{AB}=\mathrm{DE}$ (see Fig. 7.12).

Now what do you observe? You may observe that

$$ \begin{aligned} \mathrm{AB} & =\mathrm{DE} \quad \text {(Assumed)}\\ \angle \mathrm{B} & =\angle \mathrm{E} \quad \text {(Given)}\\ \mathrm{BC} & =\mathrm{EF} \quad \text {(Given)} \\ \Delta \mathrm{ABC} & \cong \Delta \mathrm{DEF} \quad \text {(By SAS rule)} \end{aligned} $$

So,

Fig. 7.12

**Case (ii) :** Let if possible $A B>D E$. So, we can take a point $P$ on $A B$ such that $\mathrm{PB}=\mathrm{DE}$. Now consider $\triangle \mathrm{PBC}$ and $\triangle \mathrm{DEF}$ (see Fig. 7.13).

Fig. 7.13

Observe that in $\triangle \mathrm{PBC}$ and $\triangle \mathrm{DEF}$,

$$ \begin{aligned} & \mathrm{PB}=\mathrm{DE} \quad \text {(By construction)} \\ & \angle \mathrm{B}=\angle \mathrm{E} \quad \text {(Given)} \\ & \mathrm{BC}=\mathrm{EF} \quad \text {(Given)} \end{aligned} $$

So, we can conclude that:

$\Delta \mathrm{PBC} \cong \triangle \mathrm{DEF}$, by the SAS axiom for congruence.

Since the triangles are congruent, their corresponding parts will be equal.

So, $\quad \angle \mathrm{PCB}=\angle \mathrm{DFE}$

But, we are given that

$\quad\ \quad \angle \mathrm{ACB}=\angle \mathrm{DFE}$

Is this possible?

This is possible only if $\mathrm{P}$ coincides with $\mathrm{A}$.

or,$\quad\mathrm{BA}=\mathrm{ED}$

So,$\quad \Delta \mathrm{ABC} \cong \Delta \mathrm{DEF} \quad \text {(by SAS axiom)}$

**Case (iii) :** If $\mathrm{AB}<\mathrm{DE}$, we can choose a point $\mathrm{M}$ on $\mathrm{DE}$ such that $\mathrm{ME}=\mathrm{AB}$ and repeating the arguments as given in Case (ii), we can conclude that $\mathrm{AB}=\mathrm{DE}$ and so, $\triangle \mathrm{ABC} \cong \triangle \mathrm{DEF}$.

Suppose, now in two triangles two pairs of angles and one pair of corresponding sides are equal but the side is not included between the corresponding equal pairs of angles. Are the triangles still congruent? You will observe that they are congruent. Can you reason out why?

You know that the sum of the three angles of a triangle is $180^{\circ}$. So if two pairs of angles are equal, the third pair is also equal $\left(180^{\circ}-\right.$ sum of equal angles $)$.

So, two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. We may call it as the AAS Congruence Rule.

Now let us perform the following activity :

Draw triangles with angles $40^{\circ}, 50^{\circ}$ and $90^{\circ}$. How many such triangles can you draw?

In fact, you can draw as many triangles as you want with different lengths of sides (see Fig. 7.14).

Fig. 7.14

Observe that the triangles may or may not be congruent to each other.

So, equality of three angles is not sufficient for congruence of triangles. Therefore, for congruence of triangles out of three equal parts, one has to be a side.

Let us now take some more examples.

**Example 3 :** Line-segment $\mathrm{AB}$ is parallel to another line-segment $\mathrm{CD}$. $\mathrm{O}$ is the mid-point of $A D$ (see Fig. 7.15). Show that (i) $\triangle A O B \cong \triangle D O C$ (ii) $O$ is also the mid-point of $\mathrm{BC}$.

**Solution :** (i) Consider $\triangle \mathrm{AOB}$ and $\triangle \mathrm{DOC}$.

$\quad \angle \mathrm{ABO}=\angle \mathrm{DCO}\quad$(Alternate angles as $\mathrm{AB} || \mathrm{CD}$ and $\mathrm{BC}$ is the transversal)

$\quad \angle \mathrm{AOB}=\angle \mathrm{DOC}\quad$(Vertically opposite angles)

$\quad \mathrm{OA}=\mathrm{OD}\quad$ (Given)

Fig. 7.15

Therefore, $\quad \triangle \mathrm{AOB} \cong \triangle \mathrm{DOC} \quad$ (AAS rule)

(ii) $\quad \mathrm{OB}=\mathrm{OC}\quad$(CPCT)

So, $\mathrm{O}$ is the mid-point of $\mathrm{BC}$.

### EXERCISE 7.1

**1.** In quadrilateral $\mathrm{ACBD}$,
$\mathrm{AC}=\mathrm{AD}$ and $\mathrm{AB}$ bisects $\angle \mathrm{A}$ (see Fig. 7.16). Show that $\triangle \mathrm{ABC} \cong \triangle \mathrm{ABD}$. What can you say about $\mathrm{BC}$ and $\mathrm{BD}$ ?

Fig. 7.16

## Show Answer

**Solution**

$\triangle A B C \quad \triangle A B D$. What can you say about $B C$ and $B D$ ?

In $\triangle A B C$ and $\triangle A B D$,

$A C=A D$ (Given)

$\angle CAB=\angle DAB(AB$ bisects $\angle A)$

$A B=A B$ (Common)

$\therefore \quad \triangle ABC \cong \triangle ABD$ (By SAS congruence rule)

$\therefore \quad BC=BD(By CPCT)$

Therefore, BC and BD are of equal lengths.

**2.** $A B C D$ is a quadrilateral in which $A D=B C$ and $\angle \mathrm{DAB}=\angle \mathrm{CBA}$ (see Fig. 7.17). Prove that

(i) $\triangle \mathrm{ABD} \cong \triangle \mathrm{BAC}$

(ii) $\mathrm{BD}=\mathrm{AC}$

(iii) $\angle \mathrm{ABD}=\angle \mathrm{BAC}$.

## Show Answer

**Solution**

In $\triangle A B D$ and $\triangle B A C$,

$A D=B C($ Given)

$\angle \angle$

$DAB=CBA($ Given)

$AB=BA$ (Common)

$\therefore \triangle ABD \cong \triangle BAC$ (By SAS congruence rule)

$\therefore BD=AC$ (By CPCT) And, $\angle ABD$

$=\widehat{B A C}(B y C P C T)$

**3.** $\mathrm{AD}$ and $\mathrm{BC}$ are equal perpendiculars to a line segment $A B$ (see Fig. 7.18). Show that CD bisects $\mathrm{AB}$.

## Show Answer

**Solution**

In $\triangle B O C$ and $\triangle A O D$,

$ \begin{aligned} & \angle \quad \angle BOC=AOD \text{ (Vertically opposite angles) } \\ & \angle \quad CBO=DAO(\text{ Each } 90^{\circ}) \\ & BC=AD \text{ (Given) } \\ & \therefore \quad \triangle BOC \cong \triangle AOD \text{ (AAS congruence rule) } \\ & \therefore \quad BO=AO(By C P C T) \\ & \Rightarrow \quad CD \text{ bisects } AB . \end{aligned} $

**4.** $\quad l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig. 7.19). Show that $\triangle \mathrm{ABC} \cong \triangle \mathrm{CDA}$.

Fig. 7.19

## Show Answer

**Solution**

In $\triangle ABC$ and $\triangle CDA$,

$ \begin{aligned} & \angle BAC=\angle DCA \text{ (Alternate interior angles, as } p | q) \\ & AC=CA \text{ (Common) } \\ & \angle \quad \angle BCA=DAC \text{ (Alternate interior angles, as } \mid | m \text{ ) } \\ & \therefore \quad \therefore \quad \triangle ABC \triangle CDA(By \text{ ASA congruence rule) } \\ & \text{ Question 5: } \end{aligned} $

**5.** Line $l$ is the bisector of an angle $\angle \mathrm{A}$ and $\mathrm{B}$ is any point on $l$. $\mathrm{BP}$ and $\mathrm{BQ}$ are perpendiculars from $\mathrm{B}$ to the arms of $\angle A$ (see Fig. 7.20). Show that:

(i) $\triangle \mathrm{APB} \cong \triangle \mathrm{AQB}$

(ii) $\mathrm{BP}=\mathrm{BQ}$ or $\mathrm{B}$ is equidistant from the arms of $\angle \mathrm{A}$.

Fig. 7.20

## Show Answer

**Solution**

In $\triangle A P B$ and $\triangle A Q B$,

$ \begin{aligned} & \therefore \quad \therefore \quad APB=AQB(\text{ Each 90) } \\ & \therefore \quad \therefore \quad PAB=QAB(I \text{ is the angle bisector of } \therefore A) \end{aligned} $

$A B=A B$ (Common)

$\therefore \triangle APB \therefore \triangle AQB$ (By AAS congruence rule) $\therefore BP=$

BQ (By CPCT)

rms of $\therefore A$. Or, it can be said that $B$ is equidistant from the a

**6.** In Fig. 7.21, $\mathrm{AC}=\mathrm{AE}, \mathrm{AB}=\mathrm{AD}$ and $\angle \mathrm{BAD}=\angle \mathrm{EAC}$. Show that $\mathrm{BC}=\mathrm{DE}$.

Fig. 7.21

## Show Answer

**Solution**

It is given that $\therefore B A D=\therefore E A C$

$\therefore BAD+\therefore DAC=\therefore EAC+\therefore DAC$

$\therefore B A C=\therefore DAE$

In $\triangle B A C$ and $\triangle D A E, A B=A D$

(Given) $\therefore BAC=$

$\therefore$ DAE (Proved above)

$A C=A E($ Given)

$\therefore \quad \triangle BAC \therefore \triangle DAE$ (By SAS congruence rule)

$\therefore \quad BC=DE(By CPCT)$

**7.** $\mathrm{AB}$ is a line segment and $\mathrm{P}$ is its mid-point. $\mathrm{D}$ and $\mathrm{E}$ are points on the same side of $\mathrm{AB}$ such that $\angle \mathrm{BAD}=\angle \mathrm{ABE}$ and $\angle \mathrm{EPA}=\angle \mathrm{DPB}$ (see Fig. 7.22). Show that

(i) $\triangle \mathrm{DAP} \cong \triangle \mathrm{EBP}$

(ii) $\mathrm{AD}=\mathrm{BE}$

Fig. 7.22

## Show Answer

**Solution**

It is given that EPA $=$ DPB

$\therefore \quad \therefore E P A+D ’ P E=D P B+D P E$

$\therefore \therefore$ DPA $=$ ÉPB

In $\sqrt{\Delta} \sqrt{\Delta}$ EBP,

$\therefore$ DAP $=$ 岸 (Given)

$A P=B P(P$ is mid-point of $A B)$

$\therefore \quad \therefore \quad$ DPA $=$ EPB (From above)

$\therefore \quad \therefore \quad \triangle$ DAP $\triangle$ EBP (ASA congruence rule)

$\therefore \quad AD=BE(By CPCT)$

**8.** In right triangle $A B C$, right angled at $C, M$ is the mid-point of hypotenuse AB. C is joined to $\mathrm{M}$ and produced to a point $\mathrm{D}$ such that $\mathrm{DM}=\mathrm{CM}$. Point $\mathrm{D}$ is joined to point $\mathrm{B}$ (see Fig. 7.23). Show that:

(i) $\triangle \mathrm{AMC} \cong \triangle \mathrm{BMD}$

(ii) $\angle \mathrm{DBC}$ is a right angle.

(iii) $\Delta \mathrm{DBC} \cong \triangle \mathrm{ACB}$

(iv) $\mathrm{CM}=\frac{1}{2} \mathrm{AB}$

Fig. 7.23

## Show Answer

**Solution**

(i) In $\triangle A M C$ and $\triangle B M D$,

$A M=B M$ ( $M$ is the mid-point of $A B$ )

$\therefore$ AMC $=\therefore$ BMD (Vertically opposite angles)

$CM=DM$ (Given)

$\therefore \quad \triangle AMC \therefore \triangle BMD$ (By SAS congruence rule)

$\therefore \quad AC=BD(By C P C T)$ And,

$\therefore$ ACM $=: BDM(By$ CPCT) ii)

$\therefore \quad ACM=` BDM($

However, AेंCM and ¿BDM are alternate interior angles.

Since alternate angles are equal,

It can be said that $D B | A C$

$ \begin{aligned} & \therefore \quad DBC+\therefore ACB=180^{\circ} \text{ (Co-interior angles) } \\ & \therefore \quad \therefore \quad DBC+90^{\circ}=180^{\circ} \\ & \therefore \quad \therefore \quad DBC=90^{\circ} \end{aligned} $

(iii) In $\triangle DBC$ and $\triangle ACB$,

$DB=AC$ (Already proved)

$\therefore DBC=\therefore ACB(.$ Each 90 $.{ }^{\circ})$

$BC=CB$ (Common)

$\therefore \triangle DBC \quad \triangle ACB$ (SAS congruence rule) iv)

$\triangle DBC \triangle ACB($

$\therefore \quad AB=DC(By CPCT)$

$\therefore \quad AB=2 CM$

$\therefore CM=\frac{1}{2} AB$

### 7.4 Some Properties of a Triangle

In the above section you have studied two criteria for congruence of triangles. Let us now apply these results to study some properties related to a triangle whose two sides are equal.

Perform the activity given below:

Construct a triangle in which two sides are equal, say each equal to $3.5 \mathrm{~cm}$ and the third side equal to $5 \mathrm{~cm}$ (see Fig. 7.24). You have done such constructions in earlier classes.

Fig. 7.24

Do you remember what is such a triangle called?

A triangle in which two sides are equal is called an isosceles triangle. So, $\triangle \mathrm{ABC}$ of Fig. 7.24 is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}$.

Now, measure $\angle \mathrm{B}$ and $\angle \mathrm{C}$. What do you observe?

Repeat this activity with other isosceles triangles with different sides. You may observe that in each such triangle, the angles opposite to the equal sides are equal.

This is a very important result and is indeed true for any isosceles triangle. It can be proved as shown below.

**Theorem 7.2 :** Angles opposite to equal sides of an isosceles triangle are equal. This result can be proved in many ways. One of the proofs is given here.

**Proof :** We are given an isosceles triangle $\mathrm{ABC}$ in which $\mathrm{AB}=\mathrm{AC}$. We need to prove that $\angle \mathrm{B}=\angle \mathrm{C}$.
Let us draw the bisector of $\angle \mathrm{A}$ and let $\mathrm{D}$ be the point of intersection of this bisector of $\angle \mathrm{A}$ and $\mathrm{BC}$ (see Fig. 7.25).In $\Delta \mathrm{BAD}$ and $\Delta \mathrm{CAD}$,

Fig. 7.25

$$ \begin{aligned} \mathrm{AB} & =\mathrm{AC} \quad \text {(Given)}\\ \angle \mathrm{BAD} & =\angle \mathrm{CAD} \quad \text {(By construction)} \\ \mathrm{AD} & =\mathrm{AD} \quad \text {(Common)} \end{aligned} $$ $$ \text { So, } \quad \Delta \mathrm{BAD} \cong \Delta \mathrm{CAD} \quad \text {(By SAS rule)} $$

So, $\angle \mathrm{ABD}=\angle \mathrm{ACD}$, since they are corresponding angles of congruent triangles.

So,$\quad \angle \mathrm{B}=\angle \mathrm{C}$

Is the converse also true? That is:

If two angles of any triangle are equal, can we conclude that the sides opposite to them are also equal?

Perform the following activity.

Construct a triangle $\mathrm{ABC}$ with $\mathrm{BC}$ of any length and $\angle \mathrm{B}=\angle \mathrm{C}=50^{\circ}$. Draw the bisector of $\angle \mathrm{A}$ and let it intersect $\mathrm{BC}$ at $\mathrm{D}$ (see Fig. 7.26).

Fig. 7.26

Cut out the triangle from the sheet of paper and fold it along $\mathrm{AD}$ so that vertex $\mathrm{C}$ falls on vertex B.

What can you say about sides $\mathrm{AC}$ and $\mathrm{AB}$ ?

Observe that $\mathrm{AC}$ covers $\mathrm{AB}$ completely

So,$\quad\mathrm{AC}=\mathrm{AB}$

Repeat this activity with some more triangles. Each time you will observe that the sides opposite to equal angles are equal. So we have the following:

**Theorem 7.3 :** The sides opposite to equal angles of a triangle are equal.

This is the converse of Theorem 7.2.

You can prove this theorem by ASA congruence rule. Let us take some examples to apply these results.

**Example 4 :** In $\triangle \mathrm{ABC}$, the bisector $\mathrm{AD}$ of $\angle \mathrm{A}$ is perpendicular to side $\mathrm{BC}$ (see Fig. 7.27). Show that $A B=A C$ and $\triangle A B C$ is isosceles.

Fig. 7.27

**Solution :** In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$,

$$ \begin{aligned} \angle \mathrm{BAD} & =\angle \mathrm{CAD}\quad \text {(Given)} \\ \mathrm{AD} & =\mathrm{AD}\quad \text {(Common)} \\ \angle \mathrm{ADB} & =\angle \mathrm{ADC}=90^{\circ}\quad \text {(Given)} \end{aligned} $$

So, $\quad \quad \triangle \mathrm{ABD} \cong \triangle \mathrm{ACD}\quad \text { (ASA rule)}$ So, $\quad \quad \mathrm{AB}=\mathrm{AC}\quad \text {(CPCT)}$ or,

$\triangle \mathrm{ABC}$ is an isosceles triangle.

**Example 5:** $\mathrm{E}$ and $\mathrm{F}$ are respectively the mid-points of equal sides $\mathrm{AB}$ and $\mathrm{AC}$ of $\triangle \mathrm{ABC}$ (see Fig. 7.28). Show that $\mathrm{BF}=\mathrm{CE}$.

Fig. 7.28

**Solution :** In $\triangle \mathrm{ABF}$ and $\triangle \mathrm{ACE}$,

$$ \begin{aligned} & \mathrm{AB}=\mathrm{AC}\quad \text {Given} \\ & \angle \mathrm{A}=\angle \mathrm{A}\quad \text {Common} \\ & \mathrm{AF}=\mathrm{AE} \quad \text { (Halves of equal sides) } \end{aligned} $$ So,$\quad \quad \Delta \mathrm{ABF} \cong \Delta \mathrm{ACE} \quad(\mathrm{SAS} \text { rule })$ Therefore, $\quad \mathrm{BF}=\mathrm{CE}\quad \text {(CPCT)}$

**Example 6 :** In an isosceles triangle $\mathrm{ABC}$ with $\mathrm{AB}=\mathrm{AC}, \mathrm{D}$ and $\mathrm{E}$ are points on $\mathrm{BC}$ such that $B E=C D$ (see Fig. 7.29). Show that $A D=A E$.

Fig. 7.29

**Solution :** In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACE}$,

$$ \begin{aligned} \mathrm{AB} & =\mathrm{AC}\quad \text {(Given)} \quad (1)\\ \angle \mathrm{B} & =\angle \mathrm{C} \end{aligned} \text {(Angles opposite to equal sides)}\quad (2) $$

Also, $\quad \mathrm{BE}=\mathrm{CD}$

So, $\quad \mathrm{BE}-\mathrm{DE}=\mathrm{CD}-\mathrm{DE}$

That is, $\quad \mathrm{BD}=\mathrm{CE}\quad(3)$

So, $\quad \Delta \mathrm{ABD} \cong \triangle \mathrm{ACE}$ (Using (1), (2), (3) and SAS rule).

This gives $\quad \mathrm{AD}=\mathrm{AE}$ $\quad(\mathrm{CPCT})$

### EXERCISE 7.2

**1.** In an isosceles triangle $\mathrm{ABC}$, with $\mathrm{AB}=\mathrm{AC}$, the bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ intersect each other at $\mathrm{O}$. Join $\mathrm{A}$ to $\mathrm{O}$. Show that :

(i) $\mathrm{OB}=\mathrm{OC}$

(ii) $\mathrm{AO}$ bisects $\angle \mathrm{A}$

## Show Answer

**Solution**

(i) It is given that in triangle $A B C, A B=A C$

$\Rightarrow \angle A C B=\angle A B C$ (Angles opposite to equal sides of a triangle are equal)

$ \begin{aligned} & \Rightarrow \frac{1}{2} \angle A C B=\frac{1}{2} \angle A B C \ & \Rightarrow \angle O C B=\angle O B C \end{aligned} $

$\Rightarrow \mathrm{OB}=\mathrm{OC}$ (Sides opposite to equal angles of a triangle are also equal)

(ii) In $\triangle O A B$ and $\triangle O A C$,

$\mathrm{AO}=\mathrm{AO}$ (Common)

$A B=A C($ Given $)$

$O B=O C$ (Proved above)

Therefore, $\triangle O A B \cong \triangle O A C$ (By SSS congruence rule)

$\Rightarrow \angle B A O=\angle C A O$ (CPCT)

$\Rightarrow \mathrm{AO}$ bisects $\angle \mathrm{A}$.

**2.** In $\triangle \mathrm{ABC}, \mathrm{AD}$ is the perpendicular bisector of $\mathrm{BC}$ (see Fig. 7.30). Show that $\triangle A B C$ is an isosceles triangle in which $\mathrm{AB}=\mathrm{AC}$.

Fig. 7.30

## Show Answer

**Solution**

In $\triangle A D C$ and $\triangle A D B$,

$A D=A D$ (Common)

$\therefore A D C=\therefore A D B($ Each 90) $C D=B D(A D$ is the perpendicular bisector of $B C)$

$\therefore \quad \triangle ADC \therefore \triangle ADB$ (By SAS congruence rule)

$\therefore \quad AB=AC(By CPCT)$

Therefore, $A B C$ is an isosceles triangle in which $A B=A C$.

**3.** $\mathrm{ABC}$ is an isosceles triangle in which altitudes $\mathrm{BE}$ and $\mathrm{CF}$ are drawn to equal sides $\mathrm{AC}$ and $\mathrm{AB}$ respectively (see Fig. 7.31). Show that these altitudes are equal.

Fig. 7.32

## Show Answer

**Solution**

In $\triangle A E B$ and $\triangle A F C$,

$\therefore A E B$ and $A F C(.$ Each $.90^{\circ}) \quad A=$

$\therefore A$ (Common angle) $A B=A C$ (Given)

$\therefore \triangle AEB \therefore \triangle AFC$ (By AAS congruence rule) $\therefore BE=$ $CF$ (By CPCT)

**4.** $\mathrm{ABC}$ is a triangle in which altitudes $\mathrm{BE}$ and $\mathrm{CF}$ to sides $A C$ and $A B$ are equal (see Fig. 7.32). Show that

(i) $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}$

(ii) $\mathrm{AB}=\mathrm{AC}$, i.e., $\mathrm{ABC}$ is an isosceles triangle.

Fig. 7.32

## Show Answer

**Solution**

(i) In $\triangle A B E$ and $\triangle A C F$,

$\therefore ABE$ and $ACF(.$ Each $90^{\circ}$ )

$\therefore A=A^{\prime}$ (Common angle)

$B E=C F$ (Given)

$\therefore \triangle ABE \therefore \triangle ACF$ (By AAS congruence rule)

(ii) It has already been proved that

$\triangle ABE \bullet \triangle ACF$

$\therefore A B=A C(B y C P C T)$

**5.** $\mathrm{ABC}$ and $\mathrm{DBC}$ are two isosceles triangles on the same base BC (see Fig. 7.33). Show that $\angle \mathrm{ABD}=\angle \mathrm{ACD}$.

Fig. 7.33

## Show Answer

**Solution**

Let us join $A D$.

In $\triangle A B D$ and $\triangle A C D$,

$A B=A C$ (Given)

$B D=C D$ (Given)

$A D=A D$ (Common side)

$\therefore \triangle ABD \cong \quad \triangle ACD$ (By SSS congruence rule)

$\therefore \quad \therefore ABD=A ACD(By CPCT)$

**6.** $\triangle \mathrm{ABC}$ is an isosceles triangle in which $\mathrm{AB}=\mathrm{AC}$. Side $\mathrm{BA}$ is produced to $\mathrm{D}$ such that $\mathrm{AD}=\mathrm{AB}$ (see Fig. 7.34). Show that $\angle B C D$ is a right angle.

Fig. 7.34

## Show Answer

**Solution**

In $\triangle ABC$,

$A B=A C$ (Given)

$\therefore \therefore ACB=\therefore ABC$ (Angles opposite to equal sides of a triangle are also equal)

In $\triangle ACD$,

$A C=A D$

$\therefore \therefore$ ADC $=: A C D$ (Angles opposite to equal sides of a triangle are also equal)

In $\triangle B C D$,

$\therefore A B C+B C D+A D C=180^{\circ}$ (Angle sum property of a triangle)

$\therefore A \dot{C B}+ACB+ACD+\therefore ACD=180^{\circ}$

$\therefore \quad \therefore 2(ACB+ACD)=180^{\circ}$

$\therefore \quad \therefore \quad 2(BCD)=180^{\circ}$

$\therefore B C D=90^{\circ}$

**7.** $\mathrm{ABC}$ is a right angled triangle in which $\angle \mathrm{A}=90^{\circ}$ and $\mathrm{AB}=\mathrm{AC}$. Find $\angle \mathrm{B}$ and $\angle \mathrm{C}$.

## Show Answer

**Solution**

It is given that

$A B=A C$

$\therefore \dot{C}=B$ (Angles opposite to equal sides are also equal)

In $\triangle ABC$

$\therefore A+\dot{B}+C=180^{\circ}$ (Angle sum property of a triangle)

$\therefore 90^{\circ}+\stackrel{\therefore}{B}+C \stackrel{\therefore}{=} 180^{\circ}$

$\therefore 90^{\circ}+\stackrel{\therefore}{B}+B \stackrel{\therefore}{=} 180^{\circ}$

$\therefore \quad \therefore \quad 90^{\circ}$

$\therefore$ :

$B=450$

$B=C=45^{\circ}$

**8.** Show that the angles of an equilateral triangle are $60^{\circ}$ each.

## Show Answer

**Solution**

Let us consider that $A B C$ is an equilateral triangle.

Therefore, $A B=B C=A C$

$A B=A C$

$\therefore C=B$ (Angles opposite to equal sides of a triangle are equal)

Also,

$AC=BC$

$\therefore B=A$ (Angles opposite to equal sides of a triangle are equal)

Therefore, we obtain $\therefore$ A

$=B \cdot C \therefore$

In $\triangle ABC$,

$\therefore A+B+C=180^{\circ}$

$\therefore A^{\circ}+A+A+180^{\circ}$

$\therefore 3 \dot{A}=180^{\circ}$

$\therefore \dot{A}=60^{\circ}$

$\therefore \dot{A}=B \stackrel{\Delta}{=} C=\ddot{6} 0^{\circ}$ Hence, in an equilateral triangle, all interior angles are of measure $60^{\circ}$.

### 7.5 Some More Criteria for Congruence of Triangles

You have seen earlier in this chapter that equality of three angles of one triangle to three angles of the other is not sufficient for the congruence of the two triangles. You may wonder whether equality of three sides of one triangle to three sides of another triangle is enough for congruence of the two triangles. You have already verified in earlier classes that this is indeed true.

To be sure, construct two triangles with sides $4 \mathrm{~cm}, 3.5 \mathrm{~cm}$ and $4.5 \mathrm{~cm}$ (see Fig. 7.35). Cut them out and place them on each other. What do you observe? They cover each other completely, if the equal sides are placed on each other. So, the triangles are congruent.

Fig. 7.35

Repeat this activity with some more triangles. We arrive at another rule for congruence.

**Theorem 7.4 (SSS congruence rule) :** If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

This theorem can be proved using a suitable construction.

You have already seen that in the SAS congruence rule, the pair of equal angles has to be the included angle between the pairs of corresponding pair of equal sides and if this is not so, the two triangles may not be congruent.

Perform this activity:

Construct two right angled triangles with hypotenuse equal to $5 \mathrm{~cm}$ and one side equal to $4 \mathrm{~cm}$ each (see Fig. 7.36).

Fig. 7.36

Cut them out and place one triangle over the other with equal side placed on each other. Turn the triangles, if necessary. What do you observe?

The two triangles cover each other completely and so they are congruent. Repeat this activity with other pairs of right triangles. What do you observe?

You will find that two right triangles are congruent if one pair of sides and the hypotenuse are equal. You have verified this in earlier classes. Note that, the right angle is not the included angle in this case.

So, you arrive at the following congruence rule:

**Theorem 7.5 (RHS congruence rule) :** If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Note that RHS stands for **Right angle - Hypotenuse - Side.**

Let us now take some examples.

**Example 7 :** $\mathrm{AB}$ is a line-segment. $\mathrm{P}$ and $\mathrm{Q}$ are points on opposite sides of $A B$ such that each of them is equidistant from the points A and B (see Fig. 7.37). Show that the line PQ is the perpendicular bisector of $\mathrm{AB}$.

**Solution :** You are given that $\mathrm{PA}=\mathrm{PB}$ and $\mathrm{QA}=\mathrm{QB}$ and you are to show that $\mathrm{PQ} \perp \mathrm{AB}$ and $\mathrm{PQ}$ bisects $\mathrm{AB}$. Let $\mathrm{PQ}$ intersect $\mathrm{AB}$ at $\mathrm{C}$.
Can you think of two congruent triangles in this figure?
Let us take $\Delta$ PAQ and $\Delta$ PBQ.

Fig. 7.37

In these triangles,

$\mathrm{AP}=\mathrm{BP}$ { (Given) }

$ \mathrm{AQ}=\mathrm{BQ}$ {(Given)}

$ \mathrm{PQ}=\mathrm{PQ}$ { (Common) }

So, $ \triangle \mathrm{PAQ} \cong \triangle \mathrm{PBQ}$ { (SSS rule) }

Therefore, $ \angle \mathrm{APQ}=\angle \mathrm{BPQ}$ { (CPCT). }

Now let us consider $\triangle \mathrm{PAC}$ and $\triangle \mathrm{PBC}$.

$ You have : \mathrm{AP} =\mathrm{BP} (Given)$

$ \angle \mathrm{APC} =\angle \mathrm{BPC}(\angle \mathrm{APQ}=\angle \mathrm{BPQ} $ { (proved above) }

$ \mathrm{PC} =\mathrm{PC} { (Common) } $

So, $ \Delta \mathrm{PAC} \cong \Delta \mathrm{PBC}$ (SAS rule)

Therefore,$ \mathrm{AC} =\mathrm{BC} (\mathrm{CPCT})(1) $

and $ \angle \mathrm{ACP} =\angle \mathrm{BCP} (\mathrm{CPCT})$

$ \begin{array}{lrll} \text{ Also, } & \angle \mathrm{ACP}+\angle \mathrm{BCP} & =180^{\circ} & \text { (Linear pair) } \\ \text { So, } & 2 \angle \mathrm{ACP} & =180^{\circ} & \\ \text { 0r } & \angle \mathrm{ACP}& =90^{\circ} & (2)\\ \end{array} $

From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB.

[Note that, without showing the congruence of $\triangle \mathrm{PAQ}$ and $\triangle \mathrm{PBQ}$, you cannot show that $\Delta \mathrm{PAC} \cong \Delta \mathrm{PBC}$ even though $\mathrm{AP}=\mathrm{BP}\quad \text {(Given)}$]

$\mathrm{PC}=\mathrm{PC} \quad \text {(Given)}$

and $\quad \angle \mathrm{PAC}=\angle \mathrm{PBC}$ (Angles opposite to equal sides in $\triangle \mathrm{APB})$

It is because these results give us SSA rule which is not always valid or true for congruence of triangles. Also the angle is not included between the equal pairs of sides.]

Let us take some more examples.

**Example 8 :** $\mathrm{P}$ is a point equidistant from two lines $l$ and $m$ intersecting at point A (see Fig. 7.38). Show that the line AP bisects the angle between them.

**Solution :** You are given that lines $l$ and $m$ intersect each other at A. Let PB $\perp l$, $\mathrm{PC} \perp m$. It is given that $\mathrm{PB}=\mathrm{PC}$.

You are to show that $\angle \mathrm{PAB}=\angle \mathrm{PAC}$.

Let us consider $\Delta \mathrm{PAB}$ and $\triangle \mathrm{PAC}$. In these two triangles,

$$ \begin{aligned} \mathrm{PB} & =\mathrm{PC}\quad \text {(Given)} \\ \angle \mathrm{PBA} & =\angle \mathrm{PCA}=90^{\circ}\quad \text {(Given)} \\ \mathrm{PA} & =\mathrm{PA}\quad \text {(Common)} \end{aligned} $$

Fig. 7.38

So, $\quad \triangle \mathrm{PAB} \cong \triangle \mathrm{PAC}\quad \text {(RHS rule)}$

So, $\quad \angle \mathrm{PAB}=\angle \mathrm{PAC}\quad \text {(CPCT)}$

Note that this result is the converse of the result proved in Q. 5 of Exercise 7.1.

### EXERCISE 7.3

**1.** $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DBC}$ are two isosceles triangles on the same base $\mathrm{BC}$ and vertices $\mathrm{A}$ and $\mathrm{D}$ are on the same side of $\mathrm{BC}$ (see Fig. 7.39). If $\mathrm{AD}$ is extended to intersect $\mathrm{BC}$ at $\mathrm{P}$, show that

(i) $\triangle \mathrm{ABD} \cong \triangle \mathrm{ACD}$

(ii) $\triangle \mathrm{ABP} \cong \triangle \mathrm{ACP}$

(iii) AP bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{D}$.

(iv) $\mathrm{AP}$ is the perpendicular bisector of $\mathrm{BC}$.

Fig. 7.39

## Show Answer

**Solution**

(i) In $\triangle A B D$ and $\triangle A C D$,

$ \begin{aligned} & A B=A C \ldots \text {…(since } \triangle A B C \text { is isosceles) } \\ & A D=A D \ldots \text {…(common side) } \\ & B D=D C \text {…(since } \triangle \mathrm{BDC} \text { is isosceles) } \\ & \triangle \mathrm{ABD} \cong \triangle \mathrm{ACD} \text {…..SSS test of congruence, } \\ & \therefore \angle \mathrm{BAD}=\angle \mathrm{CAD} \text { i.e. } \angle \mathrm{BAP}=\angle \mathrm{PAC} \text {…..[c.a.c.t]…..(i) } \end{aligned} $

(ii) In $\triangle A B P$ and $\triangle A C P$,

$ \begin{aligned} & A B=A C \ldots \text {…(since } \triangle A B C \text { is isosceles) } \\ & A P=A P \ldots \text {…ommon side) } \\ & \angle B A P=\angle P A C \text {….from (i) } \\ & \triangle A B P \cong \triangle A C P \text {…. SAS test of congruence } \\ & \therefore B P=P C \text {…[C.s.c.t]…..(ii) } \\ & \angle A P B=\angle A P C \text {….c.a.c.t. } \end{aligned} $

(iii) Since $\triangle A B D \cong \triangle A C D$

$ \angle \mathrm{BAD}=\angle \mathrm{CAD} \text {….from (i) } $

So, $A$ bisects $\angle A$

i.e. AP bisects $\angle A$…..(iii)

In $\triangle \mathrm{BDP}$ and $\triangle \mathrm{CDP}$,

DP $=$ DP …common side

$B P=P C$…from (ii)

$\mathrm{BD}=\mathrm{CD}$…(since $\triangle \mathrm{BDC}$ is isosceles)

$\triangle \mathrm{BDP} \cong \triangle \mathrm{CDP}$….SSS test of congruence

$\therefore \angle \mathrm{BDP}=\angle \mathrm{CDP}$….c.a.c.t.

$\therefore \mathrm{DP}$ bisects $\angle \mathrm{D}$

So, AP bisects $\angle \mathrm{D}$

From (iii) and (iv),

AP bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{D}$.

(iv) We know that

$\angle A P B+\angle A P C=180^{\circ} \ldots$ (angles in linear pair)

Also, $\angle \mathrm{APB}=\angle \mathrm{APC}$… from (ii)

$\therefore \angle \mathrm{APB}=\angle \mathrm{APC}=\frac{180^{\circ}}{2}=90^{\circ}$

$\mathrm{BP}=\mathrm{PC}$ and $\angle \mathrm{APB}=\angle \mathrm{APC}=90^{\circ}$

Hence, $\mathrm{AP}$ is perpendicular bisector of $\mathrm{BC}$.

**2.** $A D$ is an altitude of an isosceles triangle $A B C$ in which $A B=A C$. Show that

(i) $\mathrm{AD}$ bisects $\mathrm{BC}$

(ii) $\mathrm{AD}$ bisects $\angle \mathrm{A}$.

## Show Answer

**Solution**

(i) In $\triangle B A D$ and $\triangle C A D$,

$\therefore A D B=\therefore A D C$ (Each $90^{\circ}$ as AD is an altitude)

$A B=A C$ (Given)

$A D=A D$ (Common)

$\therefore \quad \triangle BAD \therefore \triangle CAD$ (By RHS Congruence rule)

$\therefore \quad BD=CD(By CPCT)$

Hence, $A D$ bisects $B C$.

(ii) Also, by CPCT,

$\therefore B A D=C A D$ Hence, $A D$

bisects A. :

**3.** Two sides $\mathrm{AB}$ and $\mathrm{BC}$ and median $\mathrm{AM}$ of one triangle $\mathrm{ABC}$ are respectively equal to sides $\mathrm{PQ}$ and $\mathrm{QR}$ and median $P N$ of $\triangle P Q R$ (see Fig. 7.40). Show that:

(i) $\triangle \mathrm{ABM} \cong \triangle \mathrm{PQN}$

(ii) $\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}$

Fig. 7.40

## Show Answer

**Solution**

(i) In $\triangle A B C, A M$ is the median to $B C$.

$ \begin{aligned} & \therefore \quad BM={ }^{\frac{1}{2}} BC \\ & \therefore \quad Q N={ }^{\frac{1}{2}} QR \end{aligned} $

However, $BC=QR$

$\therefore{ }^{\frac{1}{2}} _{BC}{ }^{\frac{1}{2}} QR$

$\therefore \quad BM=QN \ldots$ (1)

In $\triangle A B M$ and $\triangle P Q N$, In $\triangle P Q R, P N$ is the median to $Q R$.

$ \begin{aligned} & A B=P Q \text{ (Given) } \\ & BM=QN[\text{ From equation (1)] } \\ & A M=P N(\text{ Given) } \\ & \therefore \quad \therefore \quad \triangle ABM \quad \triangle PQN \text{ (SSS congruence rule) } \\ & \therefore \quad \therefore \quad A B M=P Q N(B y C P C T) \\ & \therefore \quad \therefore \quad A B C=P Q R \ldots(2) \end{aligned} $

(ii) In $\triangle A B C$ and $\triangle P Q R$,

$A B=P Q$ (Given)

$\therefore A B C=\therefore P Q R$ [From equation (2)]

$BC=QR$ (Given)

$\therefore \triangle ABC \therefore \triangle PQR$ (By SAS congruence rule)

**4.** $\mathrm{BE}$ and $\mathrm{CF}$ are two equal altitudes of a triangle $\mathrm{ABC}$. Using RHS congruence rule, prove that the triangle $\mathrm{ABC}$ is isosceles.

## Show Answer

**Solution**

In $\triangle B E C$ and $\triangle C F B$,

$\therefore BEC=\therefore CFB(.$ Each $.90^{\circ})$

$BC=CB$ (Common)

$B E=C F$ (Given)

$\therefore \triangle BEC \quad \triangle CFB$ (By RHS congruency)

$\therefore B B^{\prime} C=C B ‘(B y C P C T)$

$\therefore A B=A C$ (Sides opposite to equal angles of a triangle are equal)

Hence, $\triangle ABC$ is isosceles.

**5.** $\mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}$. Draw $\mathrm{AP} \perp \mathrm{BC}$ to show that $\angle \mathrm{B}=\angle \mathrm{C}$.

## Show Answer

**Solution**

In $\triangle A P B$ and $\triangle A P C$,

$\therefore APB=\therefore APC(.$ Each $.90^{\circ})$

$A B=A C($ Given $)$

$A P=A P($ Common $)$

$\therefore \triangle APB \quad \triangle APC$ (Using RHS congruence rule)

$\therefore B^{\prime \prime}=C$ (By using CPCT)

### 7.6 Summary

In this chapter, you have studied the following points :

**1.** Two figures are congruent, if they are of the same shape and of the same size.

**2.** Two circles of the same radii are congruent.

**3.** Two squares of the same sides are congruent.

**4.** If two triangles $\mathrm{ABC}$ and $\mathrm{PQR}$ are congruent under the correspondence $\mathrm{A} \leftrightarrow \mathrm{P}$, $\mathrm{B} \leftrightarrow \mathrm{Q}$ and $\mathrm{C} \leftrightarrow \mathrm{R}$, then symbolically, it is expressed as $\Delta \mathrm{ABC} \cong \Delta \mathrm{PQR}$.

**5.** If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent (SAS Congruence Rule).

**6.** If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent (ASA Congruence Rule).

**7.** If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent (AAS Congruence Rule).

**8.** Angles opposite to equal sides of a triangle are equal.

**9.** Sides opposite to equal angles of a triangle are equal.

**10.** Each angle of an equilateral triangle is of $60^{\circ}$.

**11.** If three sides of one triangle are equal to three sides of the other triangle, then the two triangles are congruent (SSS Congruence Rule).

**12.** If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent (RHS Congruence Rule).