
3. Let $A B C$ be a triangle with $\angle B A C=90^{\circ}$. Points $D, E$ on its hypotenuse $B C$ satisfy $C D=C A, B E=B A$. Let $F$ be such a point inside $\triangle A B C$ that $D E F$ is an isosceles right triangle with hypotenuse DE. Find $\angle B F C$. (Patrik Bak)


Solution. We show that $F$ is the incenter of $\triangle A B C$. Simple angle-chasing in $\triangle B F C$ then yields $\angle B F C=180^{\circ}-\frac{1}{2} B-\frac{1}{2} C=135^{\circ}$.

Since $B A=B E$, triangle $B A E$ is isosceles and $\angle B A E=90^{\circ}-\frac{1}{2} \angle B$. Similarly, $\angle D A C=90^{\circ}-\frac{1}{2} \angle C$ and simple angle-chasing gives $\angle D A E=45^{\circ}$ (Fig. 1). This means that $F$ is the circumcenter of $\triangle A D E$ : Indeed, it lies in the same half-plane determined by $B C$ as $A$, and satisfies both $\angle D F E=90^{\circ}=2 \cdot 45^{\circ}=2 \angle D A E$ and $F E=F D$.

Being the circumcenter of $\triangle A D E$, point $F$ lies on the perpendicular bisectors of $A E$ and $A D$ which coincide with angle bisectors of $\angle B, \angle C$. Hence we are done.

![](https://cdn.mathpix.com/cropped/2024_04_17_447c393b5df10bbb25bcg-14.jpg?height=452&width=812&top_left_y=1197&top_left_x=351)

Fig. 1

Another solution. We present another way to show that $F$ is the incenter of $\triangle A B C$. Denote by $M$ the midpoint of $D E$. Using the notation $B C=a, C A=b, A B=c$ we
compute $D E=b+c-a$ (Fig. 2) and

$$
B M=B D+D M=(a-b)+\frac{1}{2}(b+c-a)=\frac{1}{2}(a+c-b),
$$

hence $M$ is the point of contact of the incircle of $\triangle A B C$ with $B C$. Since $\triangle D E F$ is isosceles and right, we further have $M F=M D=\frac{1}{2} D E=\frac{1}{2}(b+c-a)$ which, according to a well-known formula, is the radius of an incircle of a right triangle. Thus $F$ is the incenter and we conclude as in the first solution.

![](https://cdn.mathpix.com/cropped/2024_04_17_447c393b5df10bbb25bcg-15.jpg?height=458&width=799&top_left_y=433&top_left_x=303)

Fig. 2

Remark. Once the contestant formulates a hypothesis that $F$ is the incenter of $\triangle A B C$, there are many other ways to prove it.
