So ∠BPD=∠BAP+∠ABP

∠CPD is the outer corner of the triangle APC.

So ∠CPD=∠CAP+∠ACP.

∠ABP=∠ACP

∠BPD=∠CPD

So ∠BAP=∠CAP.

AP is male marginal.

∠ABP=∠ACP

So triangle ABP is equal to triangle ACP(ASA).

So AB=AC

So BD=CD AD⊥BC (the height on the bottom of the isosceles triangle, the median line on the bottom and the bisector of the vertex coincide)