Let the diagonals $A C$ and $B D$ meet at $Q$, the diagonals $A D$ and $C E$ meet at $R$, and let the ray $A P$ meet the side $C D$ at $M$. We want to prove that $C M=M D$ holds. ! The idea is to show that $Q$ and $R$ divide $A C$ and $A D$ in the same ratio, or more precisely $$\frac{A Q}{Q C}=\frac{A R}{R D}$$ (which is equivalent to $Q R \| C D$ ). The given angle equalities imply that the triangles $A B C$, $A C D$ and $A D E$ are similar. We therefore have $$\frac{A B}{A C}=\frac{A C}{A D}=\frac{A D}{A E}$$ Since $\angle B A D=\angle B A C+\angle C A D=\angle C A D+\angle D A E=\angle C A E$, it follows from $A B / A C=$ $A D / A E$ that the triangles $A B D$ and $A C E$ are also similar. Their angle bisectors in $A$ are $A Q$ and $A R$, respectively, so that $$\frac{A B}{A C}=\frac{A Q}{A R}$$ Because $A B / A C=A C / A D$, we obtain $A Q / A R=A C / A D$, which is equivalent to (1). Now Ceva's theorem for the triangle $A C D$ yields $$\frac{A Q}{Q C} \cdot \frac{C M}{M D} \cdot \frac{D R}{R A}=1$$ In view of (1), this reduces to $C M=M D$, which completes the proof. Comment. Relation (1) immediately follows from the fact that quadrilaterals $A B C D$ and $A C D E$ are similar.