Convex pentagon with right angle Let $ABCDE$ be a convex pentagon wiht $CD=DE$ and $\angle{BCD}=\angle{DEA} = 90^{\circ}$. Point $F$ lies on $AB$ such that $\frac{AF}{AE}=\frac{BF}{BC}$. Prove that $\angle{FCE} = \angle{ADE}$ and $\angle{FEC}=\angle{BDC}$.
So what I did was extend $EA$ and $BC$ to meet at $G$ and drew circle $CDEG$. Then the problem was equivalent to if the ratio statement is true, then $CF\cap DA$ and $EF \cap DB$ lied on the circle. The diagram also reminded me of Pascal's theorem, but I don't know how to incorporate it due to the weird ratio condition. Any help would be appreciated.

 A: 
We have that $EA$ and $CB$ meet at $D'$, the antipode of $D$ in the circumcircle $\Gamma$ of $CDE$. If we define $B'$ as $DB\cap\Gamma$ and $A'$ as $DA\cap\Gamma$ we have that $A,B$ and $G=EB'\cap CA'$ are collinear by Pascal's theorem. Now we just have to prove $G\equiv F$. The angles at $A'$ and $B'$ are the same since $CD=DE$, hence by the sine theorem
$$ \frac{GA}{GB}=\frac{GA'\sin\widehat{ABB'}}{GB'\sin\widehat{A'AB}}=\frac{GE\sin\widehat{ABB'}}{GC\sin\widehat{A'AB}}=\frac{GE\sin\widehat{ABD}}{GC\sin\widehat{BAD}}=\frac{GE\cdot AD}{GC\cdot BD}$$
and $\frac{GA}{GB}=\frac{AE}{BC}$ (from which $G\equiv F$) follows from Menelaus' theorem.
A: No, you don't need this circle, this is not how the problem is supposed to be solved. Here is the actual solution.
Let $\angle \, CDE = \delta$. Perform a counterclockwise rotation around point $D$ of angle $\delta$ and let $G$ be the image of vertex $A$ under the rotation. Since $DE=DC$ the image of triangle $ADE$ under the rotation is triangle $GDC$ and therefore, $$\angle \, DCG = \angle \, DEA = 90^{\circ} = \angle \, BCD$$ implying that point $G$ lies on the line $BC$. Moreover, $EA = CG$. By assumption
$$\frac{BF}{FA} = \frac{BC}{EA} = \frac{BC}{CG}$$ yielding that $CF$ and $AG$ are parallel. Consequently, $$\angle \, BCE = \gamma = \angle \, BGA$$ Since $DE = DC$ triangle $ECD$ is isosceles, meaning that $$\angle \, ECD = \angle \, CED = \frac{1}{2}(180^{\circ} - \delta)$$
Due to the rotation around $D$ and the consequent congruence of triangles $ADE$ and $GDC$
$$\angle \, ADE = \angle \, GDC$$ and $$\angle \, GDA = \angle \, CDE = \delta$$ and since $DA = DG$ triangle $ADG$ is isosceles and so 
$$\angle \, AGD = \angle \, GAD =  \frac{1}{2}(180^{\circ} - \delta)$$ Consequently
$$\angle \,  AGD = \frac{1}{2}(180^{\circ} - \delta)= \angle \, ECD$$
Now we are ready to conclude that 
$$\angle \, FCE = \angle \, BCD - \angle \, BCF - \angle \, ECD = 90^{\circ} - \gamma - \frac{1}{2}(180^{\circ} - \delta) = \delta - \gamma$$ On the other hand triangle $GDC$ is right angled so $$\angle \, GDC = 90^{\circ} - \angle \, CGD = 90^{\circ} - \angle \, BGA - \angle \, AGD = 90^{\circ} -  \frac{1}{2}(180^{\circ} - \delta) - \gamma  = \delta - \gamma$$ Therefore
$$\angle \, ADE = \angle \, GDC = \delta-\gamma = \angle \, FCE$$
The equality between the other two angles is proved absolutely analogously. 

