A2. Let $A B C$ be a right-angled triangle with a right angle at $A$. Points $D, E$, and $F$ lie on the sides $A B, B C$, and $C A$ respectively, such that $|B D|=|B E|$ and $|C F|=|C E|$ (see the figure). What is the measure of the angle $\Varangle F E D$ in degrees?
(A) 30
(B) 37.5
(C) 45
(D) 52.5

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-02.jpg?height=271&width=417&top_left_y=918&top_left_x=1479)

(E) It is impossible to determine, as the answer depends on the measure of the angle $\Varangle C B A$.
A2. Let the angles of triangle $ABC$ be denoted as usual by $\alpha, \beta$, and $\gamma$. Then $\alpha=90^{\circ}$, and therefore $\beta+\gamma=$ $90^{\circ}$. Since triangles $DBE$ and $ECF$ are isosceles with vertices at $B$ and $C$, we have $\angle DEB=\frac{180^{\circ}-\beta}{2}=$ $90^{\circ}-\frac{\beta}{2}$ and $\angle CEF=\frac{180^{\circ}-\gamma}{2}=90^{\circ}-\frac{\gamma}{2}$. Therefore, $\angle FED=180^{\circ}-\angle CEF-\angle DEB=\frac{\beta+\gamma}{2}=45^{\circ}$.