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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.

This is an added diagram with my progress (or maybe digression)

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  • $\begingroup$ Geometric questions like this one go better with a diagram. $\endgroup$
    – DonAntonio
    Commented Oct 7, 2016 at 7:41
  • $\begingroup$ Thank you for notifying me. I have added a diagram. $\endgroup$
    – user371813
    Commented Oct 7, 2016 at 7:50
  • $\begingroup$ There is a typo: $F$ belongs to $AB$, not to $AC$. $\endgroup$ Commented Oct 7, 2016 at 7:51
  • $\begingroup$ Sorry for the error, it is fixed now. $\endgroup$
    – user371813
    Commented Oct 7, 2016 at 8:12

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much. I was wondering how the ending follows from menelaus? Which triangle do you use? and what colinear points do you use? $\endgroup$
    – user371813
    Commented Oct 7, 2016 at 10:40
  • $\begingroup$ @JaharMehru: I used the triangle $D'EC$ cut by the transversal $AB$, plus some trigonometry. There might be more elegant ways: a repeated application of Menelaus' theorem also gives a proof of Pascal's theorem, hence something might be redundant in the above lines. But a simple approach is to construct a cyclic hexagon first, then deal with the awkard (cross-)ratio constraint at last. $\endgroup$ Commented Oct 7, 2016 at 10:45
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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.

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