In a convex quadrilateral $ABCD$, $AB=BC+DA.$ Bisectors of ABC and BAD meet at $P$. Show that $CP=DP$.
Sadly, no ideas on this one.

2. Originally Posted by atreyyu
In a convex quadrilateral $ABCD$, $AB=BC+DA.$ Bisectors of ABC and BAD meet at $P$. Show that $CP=DP$.
Sadly, no ideas on this one.
1. I assume that you mean the angle bisector ... ? (Drawn in blue) If so:

2. Draw a sketch (see attachment)

3. For symmetry reasons the 2 orange segments must be as long as the red segment.