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DeltaABC and DeltaDBC are two isosceles ...

`DeltaABC` and `DeltaDBC` are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that :
(i) `DeltaABD cong DeltaACD`
(ii) `DeltaABP cong DeltaACP`
(iii) AP bisects `angleA` as well as `angleD`
(iv) AP is the perpendicular bisector of BC.

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