`DeltaA B C`and `DeltaD B C`are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (i) `\ DeltaA B D~=DeltaA C D`
(ii) `DeltaA B P~=DeltaACP`
(iii) `AP` bisects `∠ A` as well as `∠ D`
(iv) `AP` is the perpendicular bisector of `BC`

Explanation

(i) In ΔABD and ΔACD,


`AB = AC` (Equal sides of isosceles ΔABC)


`BD = CD` (Equal sides of isosceles ΔDBC)


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