Lemme 1. The circles AEF, BDF, and CDE intersect at $M$.
This immediately follows from Miquel's theorem, but let's recall the proof:
$(M F, M E) = (M F, M D) + (M D, M E) = (B F, B D) + (C D, C E) = (A B, C D) + (C D, A C) = (A B, A C)$, so $A, E, F, M$ are concyclic.

Lemme 2. The triangles $A E F$ and $A B C$ are (indirectly) similar.
The triangles $A F C$ and $A E B$ are right-angled at $F$ and $E$, and their angles at $A$ are equal, so they are similar. We deduce that $\frac{A E}{A F} = \frac{A B}{A C}$. This can be written as $\frac{A E}{A B} = \frac{A F}{A C}$. Since $\widehat{E A F} = \widehat{B A C}$, the triangles EAF and $B A C$ are similar. It is clear that the similarity is indirect since the oriented angles $(\overrightarrow{A B}, \overrightarrow{A C})$ and $(\overrightarrow{A E}, \overrightarrow{A F})$ are opposite.

Let's return to the exercise. Let $M^{\prime}$ be the other intersection point of the angle bisector of $\widehat{B A C}$ with the circle $A E F$. Similarly, define $M^{\prime \prime}$ as the other intersection point of the angle bisector of $\widehat{B A C}$ with the circle $A B C$. We will show that $M^{\prime} = M$.

$$\begin{aligned} \left(M^{\prime} F, M^{\prime} D\right) & = \left(M^{\prime} F, M^{\prime} A\right) \quad \text{since } A, M, D \text{ are collinear} \\ & = (E F, E A) \quad \text{since } A, E, F, M \text{ are concyclic} \\ & = (B A, B C) \quad \text{since } A E F \text{ and } A B C \text{ are indirectly similar} \\ & = (B F, B F). \end{aligned}$$

We deduce that $M^{\prime}$ lies on the circle BDF. Similarly, it lies on the circle CDE. Consequently, it is the intersection of the three circles BDF, CDE, and $A E F$: this proves that $M^{\prime} = M$.

Finally, since the arcs MF and ME are equal (since (AM) is the angle bisector of $\widehat{F A E}$), we have $M F = M E$.

## Another Solution.

Lemme 1. Two chords $\left[\mathrm{BB}^{\prime}\right]$ and $\left[\mathrm{CC}^{\prime}\right]$ of a circle intersect at an external point $A$. Then the triangles $A B C$ and $A C^{\prime} B^{\prime}$ are indirectly similar.
!

We have $\left(B^{\prime} C^{\prime}, B^{\prime} A\right) = \left(B^{\prime} C^{\prime}, B^{\prime} B\right) = \left(C C^{\prime}, C B\right) = (C A, C B)$ and similarly $\left(C^{\prime} A, C^{\prime} B^{\prime}\right) = (B C, B A)$.
Lemme 2. $\frac{D B}{D C} = \frac{A B}{A C}$.
By the law of sines, we have $\frac{A B}{D B} = \frac{\sin \widehat{A D B}}{\sin \frac{\widehat{B A C}}{2}} = \frac{\sin \widehat{C D A}}{\sin \frac{\widehat{B A C}}{2}} = \frac{A C}{D C}$.
Let's return to the exercise. Since AFC and AEB are right-angled at $F$ and $E$ and have the same angle at $A$, they are similar. We deduce that $\frac{A F}{A E} = \frac{A C}{A B}$, which can be written as $A F \cdot A B = A E \cdot A C$. In other words, $A$ has the same power with respect to the two circles BFD and CED. Consequently, $A$ lies on the radical axis (MD).
Furthermore, the lemma 1 implies that $A F M$ and $A D B$ are similar, so $\frac{B D}{B A} = \frac{M F}{M A}$. Similarly, $\frac{C D}{C A} = \frac{M E}{M A}$. According to lemma 2, we have $\frac{B D}{B A} = \frac{C D}{C A}$, so $M F = M E$.

## Exercises for Group A