Question 40
Let $ABCD$ be a trapezoid with $AB \parallel CD$, $AB \gt CD$, and such that $BC = CD = DA$. Points $E$ and $F$ divide $AB$ into three equal parts; $E$ is between $A$ and $F$. Lines $CF$ and $DE$ intersect at $P$. Prove that $\angle APB = \angle DAB$.