# Triangles : Exercise 7.3 (Mathematics NCERT Class 9th)

Q.1     $\Delta$ ABC and $\Delta$ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
(i) $\Delta ABD \cong \,\Delta ACD$
(ii) $\Delta ABP \cong \,\Delta ACP$
(iii) AP bisects $\angle A$ as well as $\angle D$
(iv) AP is the perpendicular bisector of BC.
Sol.

(i) In $\Delta s$ ABD and ACD, we have
AB = AC [Given]
BD = DC [Given]
Therefore by SSS criterion of congruence, we have $\Delta \,ABD \cong \Delta ACD$

(ii) In $\Delta s$ ABP and ACP , we have
AB = AC [Given]
$\angle BAP = \angle PAC$
[Since $\Delta \,ABD \cong \,\Delta \,ACD$
$\Rightarrow$ $\angle BAD = \angle DAC$   $\Rightarrow$   $\angle BAP = \angle PAC$
and    AP = AP [Common]
Therefore By SAS criterion of congruence , we have  $\Delta \,ABP \cong \,\Delta \,ACP$

(iii) Since $\Delta \,ABD \cong \,\Delta \,ACD$. Therefore , $\angle BAD = \angle DAC$
$\Rightarrow$ AD bisects $\angle A$ $\Rightarrow$ AP bisects $\angle A$ ... (1)
In $\Delta s$ BDP and CDP, we have
BD = CD                   [Given]
BP = PC [Since $\Delta \,ABP\, \cong \,\Delta ACP\,$ $\Rightarrow$ BP = PC]
and DP = DP [Common]
By SSS criterion of congruence, we have
$\Delta \,BDP\, \cong \,\Delta CDP\,$
$\Rightarrow$ $\angle BDP = \angle PDC$
$\Rightarrow$ DP bisects $\angle D$ $\Rightarrow$ AP bisects $\angle D$ ... (2)
Combining (1) and (2), we get
AP bisects $\angle A$ as well as $\angle D$

(iv) Since AP stands on BC
Therefore $\angle APB + \angle APC =$ 180º [Linear Pairs]
But $\angle APB = \angle APC$ [Proved above]
Therefore $\angle APB = \angle APC = {{{{180}^o}} \over 2} = {90^o}$
Also BP = PC [Proved above]
$\Rightarrow$ AP is perpendicular bisector of BC.

Q.2     AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC               (ii) AD bisects $\angle A$.

Sol.

AD is the altitude drawn from vertex A of an isosceles $\Delta$ ABC to the opposite base BC so that AB = AC, $\angle ADC = \angle ADB =$ 90º.
Now, in $\Delta s\,$ ADB and ADC, we have
Hyp. AB = Hyp. AC [Given]
and $\angle ADC = \angle ADB$ [Since Each = 90º]
Therefore By RHS criterion of congruence, we have $\Delta \,ADB \cong \,\Delta ADC$
$\Rightarrow$ BD = DC and $\angle BAD = \angle DAC$
[Since corresponding parts of congruent triangle are equal]
Hence, AD bisects BC, which proves (i) and AD bisects $\angle A$, which proves (ii).

Q.3     Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\Delta$ PQR (see figure). Show that :
(i) $\Delta ABM \cong \Delta PQN$
(ii) $\Delta ABC \cong \Delta PQR$

Two $\Delta s$ ABC and PQR in which AB = PQ, BC = QR and AM = PN.
Since AM and PN are median of $\Delta s$ ABC and PQR respectively.
Now, BC = QR [Given]
$\Rightarrow$ ${1 \over 2}BC = {1 \over 2}QR$ (Median divides opposite sides in two equal parts)
$\Rightarrow$ BM = QN ... (1)

Now , in $\Delta s$ ABM and PQN we have
AB = PQ [Given]
BM = QN [From (i)]
and AM = PN [Given]
Therefore By SSS criterion of congruence, we have
$\Delta \,ABM\, \cong \,\Delta PQN$, which proves (i)
$\Rightarrow$ $\angle B = \angle Q$ ... (2)
[Since, corresponding parts of congruent triangle are equal]
Now, in $\Delta s$ ABC and PQR we have
AB = PQ [Given]
$\angle B = \angle Q$ [From (2)]
BC = QR [Given]
Therefore By SAS criterion of congruence, we have
$\Delta \,ABC\, \cong \,\Delta PQR$, which proves (ii)

Q.4      BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Sol.

In $\Delta s$ BCF and CBE, we have
$\angle BFC = \angle CEB$ [Since Each = 90º]
Hyp. BC = Hyp. BC [Common]
FC = EB
Therefore By R.H.S. criterion of congruence, we have
$\Delta \,BCF\, \cong \,\Delta \,CBE$
$\Rightarrow$ $\angle FBC\, = \angle ECB$
[Since corresponding parts of congruent triangles are equal]
Now, in $\Delta$ ABC
$\angle ABC = \angle ACB$ [Since $\angle FBC = \angle ECB$]
$\Rightarrow$ AB = AC
[Since sides opposite to equal angles of a triangle are equal]
Therefore $\Delta \,ABC$ is an isosceles triangle.

Q.5      ABC is an isosceles triangle with AB = AC. Draw AP $\bot$ BC to show that $\angle B = \angle C$
Sol.

In $\Delta s$ ABP and ACP, we have
AB = AC [Given]
AP = AP [Common]
and $\angle APB = \angle APC$ [Since each = 90º]
Therefore , by R.H.S. criterion of congruence we have
$\Delta \,ABP \cong \,\Delta \,ACP$
$\Rightarrow$ $\angle B = \angle C$
[Since corresponding parts of congruent triangles are equal]