For triangle ABC, there is a point X such that B-A-X. There is any point P on bisector of exterior angle CAX. Prove that PB+PC>AB+AC.
I have tried for many hours, but i have no idea how to solve it. actually, i have shown PB>AB. This is a question in 9th grade level, so please give solution by use of triangle congruency and triangle inequality.. Please don't use simillarity.
Hint: Construct point $C'$ on $BA$ extended, such that $AC= AC'$.
Prove that $AP$ is the perpendicular bisector of $CC'$. Hence $PC=PC'$.
Hint: Apply triangle inequality to triangle $PBC'$.