Let $A B C$ be an isosceles triangle at $B$, and let $F$ be a point on the bisector of $\widehat{A B C}$ such that $(A F)$ is parallel to $(B C)$. Finally, let $E$ be the midpoint of $[B C]$, and let $D$ be the symmetric point of $A$ with respect to $F$. Calculate the ratio of the distances $E F / B D$.
(solved by Pierre-Andréa Silvente)

We have $A F // B C$. Therefore, $\widehat{A F B}=\widehat{F B C}$ (alternate interior angles) and $\widehat{B F C}=\widehat{F B A}$. Since $\widehat{F B A}=\widehat{F B C}$ because $(B F)$ is the bisector of $\widehat{A B C}$. Therefore, $\widehat{A F B}=\widehat{F B C}=\widehat{B F C}=\widehat{F B A}$. Similarly, $\widehat{B A C}=$ $\widehat{A C F}$ and $\widehat{B C A}=\widehat{C A F}$, and since $\widehat{B A C}=\widehat{B C A}$ because $A B C$ is isosceles, we have $\widehat{B A C}=\widehat{A C F}=\widehat{B C A}=\widehat{F A C}$. Therefore, $\widehat{A B C}=\widehat{A F C}$, and $\widehat{B A F}=\widehat{B C F}$, hence $A B C F$ is a parallelogram.

In an isosceles triangle at $B$, the bisector and the altitude from $B$ coincide, so $A C \perp B F$, thus $A B C F$ is a rhombus. Let $O$ be the intersection point of $(A C)$ and $(B F)$, $\mathcal{D}=B F$ and $d=A C, P$ the orthogonal projection of $E$ on $(B P)$. According to Thales' theorem $(P \in[B O]$ and $E \in[B C], P E // O C):$

$$
\frac{B P}{B O}=\frac{B E}{B C}=\frac{P E}{O C}
$$

Since $\frac{B E}{B C}=\frac{1}{2}$ because $E$ is the midpoint of $[B C]$, we have $\frac{B P}{B O}=\frac{1}{2}$, so $P$ is the midpoint of $[B O]$. Furthermore, $P F=$ $P O+O F=\frac{3}{4} \mathcal{D}$ (since $O F=B O$ because $A B C D$ is a rhombus). We know that $\frac{P E}{O C}=\frac{1}{2}$, so $P E=\frac{1}{4} d$. By the Pythagorean theorem in $P E F$ right-angled at $P$:

$$
E F^{2}=P E^{2}+P F^{2}=\frac{d^{2}}{16}+\frac{9 \mathcal{D}^{2}}{16}
$$

Since $E F>0, E F=\sqrt{\frac{d^{2}}{16}+\frac{9 \mathcal{D}^{2}}{16}}$.

We have $F D=A F$ (by symmetry). Let $R$ be the orthogonal projection of $D$ on $B F$, so $A O // R D, F \in[R O]$ and $F \in[A D]$. According to Thales' theorem:

$$
\frac{F A}{F D}=\frac{F O}{F R}=\frac{A O}{R D}
$$

Since $\frac{F A}{F D}=1$, we have $F O=F R$, and $A O=R D$. Therefore, $B F=B O+2 F O=\frac{3}{2} \mathcal{D}$ and $R D=\frac{d}{2}$. By the Pythagorean theorem:

$$
B D^{2}=B R^{2}+R D^{2}=\frac{d^{2}}{4}+\frac{9 \mathcal{D}^{2}}{4}
$$

Thus, since $B D>0, B D=\sqrt{\frac{d^{2}}{4}+\frac{9 D^{2}}{4}}$. Therefore,

$$
\frac{E F}{B D}=\frac{\sqrt{\frac{d^{2}}{16}+\frac{9 D^{2}}{16}}}{\sqrt{\frac{d^{2}}{4}+\frac{9 D^{2}}{4}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}
$$