Consider $L$ as the reflection of $D$ through the perpendicular bisector of $AB$, $ALDB$ is an isosceles trapezoid and $LNMA$ is cyclic, this is a well known property, it can be shown with typical application of thales parallel theorem. on the other hand $LMPA$ is trivially cyclic and we are done.

fastlikearabbit wrote: Let $\triangle {ABC} $ be isosceles triangle with $AC=BC$ . The point $D$ lies on the extension of $AC$ beyond $C$ and is that $AC>CD$. The angular bisector of $ \angle BCD $ intersects $BD$ at point $N$ and let $M$ be the midpoint of $BD$. The tangent at $M$ to the circumcircle of triangle $AMD$ intersects the side $BC$ at point $P$. Prove that points $A,P,M$ and $N$ lie on a circle. Bulgaria 2016 problem 5

Watch $DP$ as the symmedian and the rest of the solution is trivial.

Mate007 wrote: Watch $DP$ as the symmedian and the rest of the solution is trivial. Can anybody post complete solution?

Obviously $CN||AB $. Reflect $D $ over the perpendicular bisector of $AB $ to $D'$ $\implies $ $ADD'B $ is isosceles trapezium. Let $AD'$ $\cap$ $BD $ $=$ $X $ $$-1=(B ,D ;M ,X) \implies XA \cdot XD'=XB \cdot XD = XM \cdot XN \implies AD'MN \text { is cyclic} $$And, $\angle AMP=\angle ADM $ $=$ $\angle AD'P $ $\implies $ $P $ $\in $ $\odot (AD'M) $