4. Given a right triangle $ABC$ with a right angle at $A$. On the leg $AC$, a point $D$ is marked such that $AD: DC = 1: 3$. Circles $\Gamma_{1}$ and $\Gamma_{2}$ are then constructed with centers at $A$ and $C$ respectively, passing through point $D$. $\Gamma_{2}$ intersects the hypotenuse at point $E$. Circle $\Gamma_{3}$ with center $B$ and radius $BE$ intersects $\Gamma_{1}$ inside the triangle at a point $F$ such that angle $AFB$ is a right angle. Find $BC$ if $AB = 5$.

(P. D. Mulyenko)
Answer: 13.

Solution. Let $AC = x$. Then $AD = x / 4, DC = CE = 3x / 4, BE = BC - CE = \sqrt{x^2 + 25} - 3x / 4$. According to the problem, $\angle AFB = 90^\circ$, so $AF^2 + FB^2 = AB^2$, which means $AD^2 + BE^2 = 25$. Expressing everything in terms of $x$ and simplifying, we get $13x = 12 \sqrt{x^2 + 25}$. Squaring both sides, we obtain $x^2 = 144$, so $x = 12$ and

![](https://cdn.mathpix.com/cropped/2024_05_06_b963bc6b39fa3a70c9f7g-2.jpg?height=297&width=454&top_left_y=1448&top_left_x=1463)
$BC = 13$.