NCERT Solutions for Exercise 7.3 Class 9 Maths Chapter 7 - Triangles

Q1 (i) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AD=AD$ (Common)

(ii) $AB=AC$ (Isosceles triangle)

(iii) $BD=CD$ (Isosceles triangle)

Thus by SSS congruency we can conclude that :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Q1 (ii) Triangles ABC and Triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that $\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

Answer:

Consider $\mathrm{\Delta }ABP$ and $\mathrm{\Delta }ACP$ ,

(i) $AP$ is common side in both the triangles.

(ii) $\mathrm{\angle }PAB=\mathrm{\angle }PAC$ (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii) $AB=AC$ (Isosceles triangles)

Thus by SAS axiom, we can conclude that :

$\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

Q1 (iii) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }D$ .

Answer:

In the first part, we have proved that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$ .

So, by c.p.c.t. $\mathrm{\angle }PAB=\mathrm{\angle }PAC$ .

Hence AP bisects $\mathrm{\angle }A$ .

Now consider $\mathrm{\Delta }BPD$ and $\mathrm{\Delta }CPD$ ,

(i) $PD=PD$ (Common)

(ii) $BD=CD$ (Isosceles triangle)

(iii) $BP=CP$ (by c.p.c.t. from the part (b))

Thus by SSS congruency we have :

$\mathrm{\Delta }BPD\cong \mathrm{\Delta }CPD$

Hence by c.p.c.t. we have : $\mathrm{\angle }BDP=\mathrm{\angle }CDP$

or AP bisects $\mathrm{\angle }D$ .

Q1 (iv) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP is the perpendicular bisector of BC.

Answer:

In the previous part we have proved that $\mathrm{\Delta }BPD\cong \mathrm{\Delta }CPD$ .

Thus by c.p.c.t. we can say that : $\mathrm{\angle }BPD=\mathrm{\angle }CPD$

Also, $BP=CP$

SInce BC is a straight line, thus : $\mathrm{\angle }BPD+\mathrm{\angle }CPD={180}^{\circ }$

or $2\mathrm{\angle }BPD={180}^{\circ }$

or $\mathrm{\angle }BPD={90}^{\circ }$

Hence it is clear that AP is a perpendicular bisector of line BC.

Q2 (i) AD is an altitude of an isosceles triangle ABC in which $AB=AC$ . Show that AD bisects BC

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AB=AC$ (Given)

(ii) $AD=AD$ (Common in both triangles)

(iii) $\mathrm{\angle }ADB=\mathrm{\angle }ADC={90}^{\circ }$

Thus by RHS axiom we can conclude that :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Hence by c.p.c.t. we can say that : $BD=CD$ or AD bisects BC.

Q2 (ii) AD is an altitude of an isosceles triangle ABC in which $AB=AC$ . Show that AD bisects $\mathrm{\angle }A$ .

Answer:

In the previous part of the question we have proved that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Thus by c.p.c.t., we can write :

$\mathrm{\angle }BAD=\mathrm{\angle }CAD$

Hence $AD$ bisects $\mathrm{\angle }A$ .

Q3 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\mathrm{\Delta }PQR$ (see Fig). Show that:

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

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

Answer:

(i) From the figure we can say that :

$BC=QR$

or $\frac{1}{2}BC=\frac{1}{2}QR$

or $BM=QN$

Now, consider $\mathrm{\Delta }ABM$ and $\mathrm{\Delta }PQN$ ,

(a) $AM=PN$ (Given)

(b) $AB=PQ$ (Given)

(c) $BM=QN$ (Prove above)

Thus by SSS congruence rule, we can conclude that :

$\mathrm{\Delta }ABM\cong \mathrm{\Delta }PQN$

(ii) Consider $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }PQR$ :

(a) $AB=PQ$ (Given)

(b) $\mathrm{\angle }ABC=\mathrm{\angle }PQR$ (by c.p.c.t. from the above proof)

(c) $BC=QR$ (Given)

Thus by SAS congruence rule,

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$

Q4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:

Using the given conditions, consider $\mathrm{\Delta }BEC$ and $\mathrm{\Delta }CFB$ ,

(i) $\mathrm{\angle }BEC=\mathrm{\angle }CFB$ (Right angle)

(ii) $BC=BC$ (Common in both the triangles)

(iii) $BE=CF$ (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that : $\mathrm{\Delta }BEC\cong \mathrm{\Delta }CFB$

Hence by c.p.c.t., $\mathrm{\angle }B=\mathrm{\angle }C$

And thus $AB=AC$ (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

Q5 ABC is an isosceles triangle with $AB=AC$ . Draw $AP\perp BC$ to show that $\mathrm{\angle }B=\mathrm{\angle }C$ .

Answer:

Consider $\mathrm{\Delta }ABP$ and $\mathrm{\Delta }ACP$ ,

(i) $\mathrm{\angle }APB=\mathrm{\angle }APC={90}^{\circ }$ (Since it is given that AP is altitude.)

(ii) $AB=AC$ (Isosceles triangle)

(iii) $AP=AP$ (Common in both triangles)

Thus by RHS axiom we can conclude that :

$\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

Now, by c.p.c.t.we can say that :

$\mathrm{\angle }B=\mathrm{\angle }C$