# 1.(iii)  $\small \Delta ABC$ and $\small \Delta DBC$ 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 (iii) AP bisects $\small \angle A$  as well as $\small \angle D$.

S Sanket Gandhi

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

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

Hence AP bisects $\angle A$.

Now consider $\Delta BPD$  and  $\Delta CPD$,

(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   :

$\Delta BPD\ \cong \ \Delta CPD$

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

or  AP bisects $\angle D$.

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