26th Turkish Mathematical Olympiad

Let $FD\cap BP=T$ and $AT\cap BC=R$. Using Menelaus and Ceva theorems, we get $$\frac{BC}{DC}=\frac{BF}{AF}\cdot \frac{AP}{PD}=\frac{BR}{DR}.$$ This shows that the points $B$, $R$, $D$, $C$ are harmonic. The lines $AB$, $AR$, $AD$, $AC$ form a harmonic pencil and hence it follows that the points $B$, $T$, $P$, $E$ are also harmonic. In triangle $DTE$, let $U$ and $V$ be the feet of the perpendicular lines from the vertices $E$ and $T$ to the opposite sides. Since $[DP]$ is an altitude, the lines $EU$, $TV$, $DP$ are concurrent. As the points $B$, $T$, $P$, $E$ are harmonic, using Ceva theorem, we get $$\frac{EV}{VD}\cdot \frac{DU}{UT}=\frac{EP}{TP}=\frac{EB}{TB}.$$

Therefore, the points $B$, $U$, $V$ satisfy the Menelaus theorem in the triangle $TDE$ which in turn implies that the points $B$, $U$, $V$ are collinear. Since the points $T$, $U$, $V$, $E$ are concyclic, we obtain that $\angle TUB=\angle TED$ and hence $△BUD\sim △QED$. As the similarity ratio is $BD/QD=1/2$, we conclude that $UD/ED=1/2$. Since $EU\perp UD$ we get $\angle UDE=\angle FDE=60^{\circ }$ and we are done.

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Alternative solution.

As in Solution 1 it can be shown that the points $B$, $T$, $P$, $E$ are harmonic. Let $\angle FDB=\angle QDE=\alpha$, $\angle PDT=\beta$, $\angle EDP=\theta$. We have $$\frac{\sin \alpha }{\sin (\alpha +\beta +\theta )}\cdot \frac{TD}{DE}=\frac{BT}{BE}=\frac{PT}{PE}=\frac{\sin \beta }{\sin \theta }\cdot \frac{TD}{DE}.$$ It follows that $$\frac{\sin \alpha }{\sin (\alpha +\beta +\theta )}=\frac{\sin \beta }{\sin \theta }.$$ By applying trigonometric transformation formulas we get $$\begin{gathered} \sin \alpha \cdot \sin \theta =\frac{1}{2}(\cos (\alpha -\theta )-\cos (\alpha +\theta )) \\ \sin (\alpha +\beta +\theta )\cdot \sin \beta =\frac{1}{2}(\cos (\alpha +\theta )-\cos (\alpha +2\beta +\theta )). \end{gathered}$$ Therefore, we have $$\cos (\alpha +\theta )=\frac{1}{2}(\cos (\alpha -\theta )+\cos (\alpha +2\beta +\theta ))=\cos (\alpha +\beta )\cdot \cos (\beta +\theta ).$$ Since $0^{\circ }<\alpha +\beta <90^{\circ }$, $0^{\circ }<\alpha +\theta <90^{\circ }$ and $DP\perp BQ$, we get $$\frac{\cos (\alpha +\beta )}{\cos (\alpha +\theta )}=\frac{DQ}{BD}=2.$$ Finally we obtain that $\cos (\beta +\theta )=1/2$ and $\angle FDE=\beta +\theta =60^{\circ }$ and hence we are done.