, 2016

Denote the common angle in (1) by $\theta$. As $\triangle A B F \sim \triangle A C D$, we have $\frac{A B}{A C} = \frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. From $E A = E D$, we get
$$\angle A F D = \angle A B C = 90^{\circ} + \theta = 180^{\circ} - \frac{1}{2} \angle A E D.$$
Hence, $F$ lies on the circle with centre $E$ and radius $E A$. In particular, $E F = E A = E D$. As $\angle E F A = \angle E A F = 2 \theta = \angle B F C$, points $B, F, E$ are collinear.
As $\angle E D A = \angle M A D$, we have $E D \parallel A M$ and hence $E, D, X$ are collinear. As $M$ is the midpoint of $C F$ and $\angle C B F = 90^{\circ}$, we get $M F = M B$. In the isosceles triangles $E F A$ and $M F B$, we have $\angle E F A = \angle M F B$ and $A F = B F$. Therefore, they are congruent to each other. Then we have $B M = A E = X M$ and $B E = B F + F E = A F + F M = A M = E X$. This shows $\triangle E M B \cong \triangle E M X$. As $F$ and $D$ lie on $E B$ and $E X$ respectively and $E F = E D$, we know that lines $B D$ and $X F$ are symmetric with respect to $E M$. It follows that the three lines are concurrent.

From $\angle C A D = \angle E D A$, we have $A C \parallel E D$. Together with $A C \parallel E X$, we know that $E, D, X$ are collinear. Denote the common angle in (1) by $\theta$. From $\triangle A B F \sim \triangle A C D$, we get $\frac{A B}{A C} = \frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. This yields $\angle A F D = \angle A B C = 90^{\circ} + \theta$ and hence $\angle F D C = 90^{\circ}$, implying that $B C D F$ is cyclic. Let $\Gamma_{1}$ be its circumcircle.
Next, from $\triangle A B F \sim \triangle A D E$, we have $\frac{A B}{A D} = \frac{A F}{A E}$ so that $\triangle A B D \sim \triangle A F E$. Therefore,
$$\angle A F E = \angle A B D = \theta + \angle F B D = \theta + \angle F C D = 2 \theta = 180^{\circ} - \angle B F A.$$
This implies $B, F, E$ are collinear. Note that $F$ is the incentre of triangle $D A B$. Point $E$ lies on the internal angle bisector of $\angle D B A$ and lies on the perpendicular bisector of $A D$. It follows that $E$ lies on the circumcircle $\Gamma_{2}$ of triangle $A B D$, and $E A = E F = E D$.
Also, since $C F$ is a diameter of $\Gamma_{1}$ and $M$ is the midpoint of $C F$, $M$ is the centre of $\Gamma_{1}$ and hence $\angle A M D = 2 \theta = \angle A B D$. This shows $M$ lies on $\Gamma_{2}$. Next, $\angle M D X = \angle M A E = \angle D X M$ since $A M X E$ is a parallelogram. Hence $M D = M X$ and $X$ lies on $\Gamma_{1}$.

We now have two ways to complete the solution.
- Method 1. From $E F = E A = X M$ and $E X \parallel F M$, $E F M X$ is an isosceles trapezoid and is cyclic. Denote its circumcircle by $\Gamma_{3}$. Since $B D$, $E M$, $F X$ are the three radical axes of $\Gamma_{1}$, $\Gamma_{2}$, $\Gamma_{3}$, they must be concurrent.
- Method 2. As $\angle D M F = 2 \theta = \angle B F M$, we have $D M \parallel E B$. Also,
$$\angle B F D + \angle X B F = \angle B F C + \angle C F D + 90^{\circ} - \angle C B X = 2 \theta + (90^{\circ} - \theta) + 90^{\circ} - \theta = 180^{\circ}$$
implies $D F \parallel X B$. These show the corresponding sides of triangles $D M F$ and $B E X$ are parallel. By Desargues' Theorem, the two triangles are perspective and hence $D B$, $M E$, $F X$ meet at a point.

Let the common angle in (1) be $\theta$. From $\triangle A B F \sim \triangle A C D$, we have $\frac{A B}{A C} = \frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. Then $\angle A D F = \angle A C B = 90^{\circ} - 2 \theta = 90^{\circ} - \angle B A D$ and hence $D F \perp A B$. As $F A = F B$, this implies $\triangle D A B$ is isosceles with $D A = D B$. Then $F$ is the incentre of $\triangle D A B$.
Next, from $\angle A E D = 180^{\circ} - 2 \theta = 180^{\circ} - \angle D B A$, points $A, B, D, E$ are concyclic. Since we also have $E A = E D$, this shows $E, F, B$ are collinear and $E A = E F = E D$.

Note that $C$ lies on the internal angle bisector of $\angle B A D$ and lies on the external angle bisector of $\angle D B A$. It follows that it is the $A$-excentre of triangle $D A B$. As $M$ is the midpoint of $C F$, $M$ lies on the circumcircle of triangle $D A B$ and it is the centre of the circle passing through $D, F, B, C$. By symmetry, $D E F M$ is a rhombus. Then the midpoints of $A X$, $E M$ and $D F$ coincide, and it follows that $D A F X$ is a parallelogram.
Let $P$ be the intersection of $B D$ and $E M$, and $Q$ be the intersection of $A D$ and $B E$. From $\angle B A C = \angle D C A$, we know that $D C$, $A B$, $E M$ are parallel. Thus we have $\frac{D P}{P B} = \frac{C M}{M A}$. This is further equal to $\frac{A E}{B E}$ since $C M = D M = D E = A E$ and $M A = B E$. From $\triangle A E Q \sim \triangle B E A$, we find that
$$\frac{D P}{P B} = \frac{A E}{B E} = \frac{A Q}{B A} = \frac{Q F}{F B}$$
by the Angle Bisector Theorem. This implies $Q D \parallel F P$ and hence $F, P, X$ are collinear, as desired.