Triangle ABC and Triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC.If AD is extended to intersect BC at P,show that
(a)Triangle ABD ≈ Triangle ACD
(b)Triangle ABP ≈ Triangle ACP
(c)AP bisects angle a as well as angle d
(d)AP is the perpendicular bisector of BC
(a) For the congruence of $\triangle ABD$ and $\triangle ACD$, use the fact that the sides match. The criterion, in North American schools, is often called SSS.
(b) It follows from the result in (a) that $\angle BAP=\angle CAP$. Then what you need should follow from the congruence criterion often called SAS (side, angle, side).
(c) The fact that $\angle BAP=\angle CAP$, that is, the fact that $AP$ bisects $\angle BAC$, has already been mentioned and proved. Essentially the same argument works for $\angle BDC$.
(d) Enough has been proved above to make this straightforward.