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1. $\small \Delta ABC$ 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 AP bisects $\small \angle A$ as well as $\small \angle D$ .

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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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Sanket Gandhi

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