Question 77
Let $ABCD$ be a convex quadrilateral, and let $P$ be the intersection of the external bisectors of $\angle DAC$ and $\angle DBC$. Prove that $\angle APD=\angle BPC$ if and only if $AD+AC=BC+BD$.
[Note: The external bisector of an angle is the line through its vertex perpendicular to its internal (usual) bisector.]
[Note: The external bisector of an angle is the line through its vertex perpendicular to its internal (usual) bisector.]