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  • ∆ ABC and ∆ 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.

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 thatAP is the perpendicular bisector of BC.

   

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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}

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

  \angle BPD\ =\ 90^{\circ}

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

Posted by

Sanket Gandhi

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