Problem 1813

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 $\triangle BUD \sim \triangle QED$. 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.

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
$$\sin \alpha \cdot \sin \theta = \frac{1}{2}(\cos(\alpha - \theta) - \cos(\alpha + \theta)) \\ \sin(\alpha + \beta + \theta) \cdot \sin \beta = \frac{1}{2}(\cos(\alpha + \theta) - \cos(\alpha + 2\beta + \theta)).$$
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.