Question

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

Answer 1

(i) In $△ABD$ and $△ACD$,

$AB=AC$ ....(since $△ABC$ is isosceles)

$AD=AD$ ....(common side)

$BD=DC$ ....(since $△BDC$ is isosceles)

$\Delta ABD\cong \Delta ACD$ .....SSS test of congruence,

$\therefore \angle BAD=\angle CAD$ i.e. $\angle BAP=\angle PAC$ .....[c.a.c.t]......(i)

(ii) In $△ABP$ and $△ACP$,

$AB=AC$ ...(since $△ABC$ is isosceles)

$AP=AP$ ...(common side)

$\angle BAP=\angle PAC$ ....from (i)

$△ABP\cong △ACP$ .... SAS test of congruence

$\therefore BP=PC$ ...[c.s.c.t].....(ii)

$\angle APB=\angle APC$ ....c.a.c.t.

(iii) Since $△ABD\cong △ACD$

$\angle BAD=\angle CAD$ ....from (i)

So, $AD$ bisects $\angle A$

i.e. $AP$ bisects$\angle A$.....(iii)

In $△BDP$ and $△CDP$,

$DP=DP$ ...common side

$BP=PC$ ...from (ii)

$BD=CD$ ...(since $△BDC$ is isosceles)

$△BDP\cong △CDP$ ....SSS test of congruence

$\therefore \angle BDP=\angle CDP$ ....c.a.c.t.

$\therefore$ $DP$ bisects$\angle D$

So, $AP$ bisects $\angle D$ ....(iv)

From (iii) and (iv),

$AP$ bisects $\angle A$ as well as $\angle D$.

(iv) We know that

$\angle APB+\angle APC={180}^{\circ }$ ....(angles in linear pair)

Also, $\angle APB=\angle APC$ ...from (ii)

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

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

Hence, $AP$ is perpendicular bisector of $BC$.