Problem on isosceles triangle

Question

If $P$ is any point on a straight line drawn through the vertex $A$ of an isosceles triangle $ABC$, parallel to the base, prove that $PB+PC>AB+AC$

Please only give hint and tell how should I start

Answer 1

Hint: reflect $B$ with respect to $AP$. Apply triangle inequality.

Answer 2

Extend BA to C’ such that BA = AC’.

Prove that PC = PC’ by showing that AP is the perpendicular bisector of CC’.

Then, apply triangle inequality to $\triangle PBC’$.