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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.


Answers (1)


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.

Posted by

Sanket Gandhi

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