Subfield: Geometry


Question: In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with

$$
\angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A .\tag{1 }
$$

Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.

Solution:
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 $E D / / A M$ 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.

<img_3935>
From $\angle C A D=\angle E D A$, we have $A C / / E D$. Together with $A C / / E X$, 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 $\Gamma_{1}$.

<img_3322>

We now have two ways to complete the solution.

- Method 1. From $E F=E A=X M$ and $E X / / F M, E F M X$ 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 $D M / / E B$. Also,

$$
\angle B F D+\angle X B F=\angle B F C+\angle C F D+90^{\circ}-\angle C B X=2 \theta+\left(90^{\circ}-\theta\right)+90^{\circ}-\theta=180^{\circ}
$$

implies $D F / / X B$. 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 $E A=E F=E D$.

<img_3596>

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 $Q D / / F P$ and hence $F, P, X$ are collinear, as desired.