National Olympiad of Argentina

Extend $PA$ and $PB$ to meet $CD$ at $X$ and $Y$ respectively. Since $XY\parallel AB$, Thales' theorem yields $\frac{XD}{AE}=\frac{PD}{PE}=\frac{CD}{FE}$. Also $AE=FE$, so $XD=CD$, i.e. $D$ is the midpoint of $XC$. In addition $CD=DA$, hence $DA=DX=DC$. Thus triangle $XCA$ is right at $A$, so that $PA\perp AC$. By symmetry $PB\perp BD$. Let $O$ be the circumcenter of $ABCD$ (an isosceles trapezoid is cyclic). Then $O$ and $D$ are both equidistant from $A$ and $C$, hence $OD$ is the perpendicular bisector of segment $AC$. In particular $OD\perp AC$ and likewise $OC\perp BD$.



Now $PA\perp AC$, $PB\perp BD$ yield $PA\parallel OD$; likewise $PB\parallel OC$. Hence $\angle APB=\angle DOC$. Note that $A$ and $O$ are on the same side of chord $CD$ in the circumcircle, hence $\angle DOC=2\angle DAC$. Note also that $AC$ is the bisector of $\angle DAB$ because $CD=CB$ due to $CD=CB$. It follows that $\angle DAB=2\angle DAC=\angle DOC$ and $\angle APB=\angle DOC=\angle DAB$.