# Triangles : Exercise 7.2 (Mathematics NCERT Class 9th)

Q.1     In an isosceles triangle ABC, with AB = AC, the bisectors of $\angle B\,and\,\angle C$ intersect each other at O. Join A to O. Show that :
(i) OB = OC               (ii) AO bisects $\angle A$.
Sol.

(i) In $\Delta$ ABC, we have
AB = AC
$\Rightarrow$ $\angle C = \angle B$ [Since angles opposite to equal sides are equal] $\Rightarrow$ ${1 \over 2}\angle B = {1 \over 2}\angle C$
$\Rightarrow$ $\angle OBC = \angle OCB$
[Since OB and OC bisect $\angle s$ B and C respectively. Therefore $\angle OBC = {1 \over 2}\angle B\,\,and\,\,\angle OCB = {1 \over 2}\angle C$]
$\Rightarrow$ OB = OC [Since sides opp. to equal $\angle s$ are equal]
(ii) Now, in $\Delta s$ ABO and ACO , we have

AB = AC [Given]
$\angle OBC = \angle OCB$ [From (1)]
OB = OC [From (2)]
Therefore By SAS criterion of congruence, we have
$\Delta ABO \cong \Delta \,ACO$
$\Rightarrow$ $\angle BAO = \angle CAO$
[Since corresponding parts of congruent triangles are equal]
$\Rightarrow$ AO bisects $\angle A$.

Q.2      In $\Delta$ ABC, AD is the perpendicular bisector of BC (see figure). Show that $\Delta$ ABC is an isosceles triangle in which AB = AC.

In $\Delta s$ ABD and ACD, we have
DB = DC [Given]
$\angle ADB = \angle ADC$ [since AD $\bot$ BC]
Therefore by SAS criterion of congruence, we have. $\Delta \,ABD\, \cong \Delta ACD$
$\Rightarrow$ AB = AC
[Since corresponding parts of congruent triangles are equal]
Hence, $\Delta$ ABC is isosceles.

Q.3      ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.

In $\Delta s$ ABE and ACF, we have
$\angle AEB = \angle AFC$ [Since Each = 90º]
$\angle BAE = \angle CAF$ [Common]
and BE = CF [Given]
Therefore By AAS criterion of congruence, we have
$\Delta \,ABE\, \cong \,\Delta \,ACF$
$\Rightarrow$ AB = AC
[Since Corresponding parts of congruent triangles are equal]
Hence, $\Delta$ ABC is isosceles.

Q.4      ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that

(i) $\Delta \,ABE\, \cong \,\Delta \,ACF$

(ii) AB = AC , i.e., ABC is an isosceles triangle.

Let BE $\bot$ AC and CF $\bot$ AB.
In $\Delta s$ ABE and ACF, we have
$\angle AEB = \angle AFC$ [Since Each = 90º]
$\angle A = \angle A$ [Common]
and , AB = AC [Given]
Therefore By AAS criterion of congruence,
$\Delta ABE \cong \Delta ACF$
$\Rightarrow$ BE = CF
[Since corresponding parts of congruent triangles are equal]

Q.5    ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that $\angle ABD = \angle ACD$.

In $\Delta$ ABC, we have
AB = AC
$\Rightarrow$ $\angle ABC = \angle ACB$ ... (1)
[Since angles opposite to equal sides are equal]
In $\Delta \,BCD$, we have
BD = CD
$\Rightarrow$ $\angle DBC = \angle DCB$ ... (2)
[Since angles opposite to equal sides are equal]
Adding (1) and (2), we have
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow$ $\angle ABC = \angle ACD$

Q.6      $\Delta \,$ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that $\angle BCD$ is a   right angle.

In $\Delta$ ABC, we have
AB = AC    [given]
$\Rightarrow$ $\angle ACB = \angle ABC$ ... (1)
[Since angles opp. to equal sides are equal]
Therefore AD = AC [Since AB = AC]
Thus , in $\Delta$ ADC, we have
$\Rightarrow$ $\angle ACD = \angle ADC$ ... (2)
[Since angles opp. to equal sides are equal]
Adding (1) and (2) , we get
$\angle ACB + \angle ACD = \angle ABC + \angle ADC$
$\Rightarrow$ $\angle BCD = \angle ABC + \angle BDC$ [Since $\angle ADC = \angle BDC$]
$\Rightarrow$ $\angle BCD + \angle BCD = \angle ABC + \angle BDC + \angle BCD$  [Adding $\angle BCD$ on both sides]
$\Rightarrow$ $2\angle BCD = 180^\circ$ [Angle sum proprty]
$\Rightarrow$ $\angle BCD =$ 90º
Hence , $\angle BCD$ is a right angle.

Q.7     ABC is a right angled triangle in which $\angle A =$ 90º and AB = AC. Find $\angle B\,\,and\,\angle C$.
Sol.

We have,
$\angle A =$ 90º
AB = AC
$\Rightarrow$ $\angle C= \angle B$
[Since angles opp. to equal sides of a triangle are equal]
Also $\angle A + \angle B + \angle C =$ 180º [Angle - sum property]
$\Rightarrow$ 90º + $2\angle B =$ 180º [Since $\angle C = \angle B$]
$\Rightarrow$ $2\angle B =$ 180º – 90º = 90º
$\Rightarrow$ $\angle B = {{{{90}^o}} \over 2} = {45^o}$
Therefore $\angle C = \angle B = {45^o}$

Q.8      Show that the angles of an equilateral triangle are 60º each.
Sol.

Let $\Delta$ ABC be an equilateral triangle so that AB = AC = BC.
Now Since AB = AC
$\Rightarrow$ $\angle C = \angle B$ ... (1)  [Since angles opp. to equal sides are equal]
Also since, CB = CA
$\Rightarrow$ $\angle A = \angle B$ .... (2)  [Since angles opp. to equal sides are equal]
From (1) and (2), we have $\angle A = \angle B = \angle C$
Also $\angle A + \angle B + \angle C =$ 180º [Angle - sum property]
Thererfore $\angle A + \angle A + \angle A =$ 180º
$\Rightarrow$ $3\angle A =$ 180º
$\Rightarrow$$\angle A =$${{180^\circ }\over 3}$
$\Rightarrow$$\angle A =$ 60º

Therefore $\angle A = \angle B = \angle C =$ 60º
Thus , each angle of an equilateral triangle is 60º.