1.(iv) \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

(iv) AP is the perpendicular bisector of BC.

               

Answers (1)
S Sanket Gandhi

In the previous part we have proved that   \Delta BPD\ \cong \ \Delta CPD.

Thus by c.p.c.t. we can say that  :       \angle BPD\ =\ \angle CPD

Also,                                                         BP\ =\ CP

SInce BC is a straight line, thus  :           \angle BPD\ +\ \angle CPD\ =\ 180^{\circ}

or                                                                                    2\angle BPD\ =\ 180^{\circ}

or                                                                                       \angle BPD\ =\ 90^{\circ}

Hence it is clear that  AP is a perpendicular bisector of line BC.

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