The question is: "$ABCD$ is a quadrilateral in which angle $B =$ angle $C$ and $AC$ bisects angle $BAD$. If $BA$ and $CD$, when extended, meet at $E$, prove that $AD/DC = AE/BE$."
I'm finding this one trickier than expected... I'd be grateful for any help. Thanks!
Let us just try to draw the figure. Since $AC$ bisects $\widehat{BAD}$, given that $B'$ is the symmetric of $B$ with respect to $AC$, we must have that $D$ lies on the $AB'$ line. Given that $l$ is the perpendicular bisector of $BC$ and $A'$ is the symmetric of $A$ with respect to $l$, in order to have $\widehat{ABC}=\widehat{BCD}$ we must have that $D$ lies on the $CA'$ line, so $$ D = AB'\cap CA'.$$ and $E\in l$.
In this configuration we have: $$\frac{AE}{BE}=\frac{AE}{EC}=\frac{AD}{DC}$$ in virtue of the bisector theorem, giving $\frac{AD}{AE}=\frac{DC}{CE}$.