Problem 10.6. In a right triangle $ABC$ with a right angle at $A$, the altitude $AH$ is drawn. A circle passing through points $A$ and $H$ intersects the legs $AB$ and $AC$ at points $X$ and $Y$ respectively. Find the length of segment $AC$, given that $AX=5$, $AY=6$, and $AB=9$.
Answer: 13.5.

Solution. Let $K$ be the second intersection point of the given circle with side $BC$ (Fig. 9). Since $\angle AHK=90^{\circ}$, segment $AK$ is the diameter of this circle, and $\angle AXK = \angle AYK = 90^{\circ}$. Therefore, $AXKY$ is a rectangle, and $XK = AY = 6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2fdcf97aa9799d0d4cd6g-38.jpg?height=422&width=497&top_left_y=92&top_left_x=476)

Fig. 9: to the solution of problem 10.6

From the condition, it follows that $XB = AB - AX = 9 - 5 = 4$. Note that right triangles $XBK$ and $ABC$ with a common acute angle $B$ are similar. Therefore,

$$
\frac{4}{6} = \frac{XB}{XK} = \frac{AB}{AC} = \frac{9}{AC}
$$

from which we find $AC = 13.5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2fdcf97aa9799d0d4cd6g-38.jpg?height=423&width=493&top_left_y=929&top_left_x=478)

Fig. 10: to the solution of problem 10.6

Another solution. From the inscribed quadrilateral $AXHY$, it follows that $\angle XHY = 180^{\circ} - \angle XAY = 90^{\circ}$. Note that $\angle CHY = 180^{\circ} - \angle XHY - \angle BHX = 90^{\circ} - \angle BHX = \angle AHX$ (Fig. 10). This means that in similar right triangles $AHC$ and $BHA$, points $Y$ and $X$ correspond to each other, that is, they divide the corresponding sides $AC$ and $AB$ in the same ratio (this can also be derived from the similarity of triangles $CHY$ and $AHX$). Therefore,

$$
\frac{6}{CY} = \frac{AY}{CY} = \frac{BX}{AX} = \frac{4}{5}
$$

from which we find $CY = 6 \cdot \frac{5}{4} = 7.5$ and $AC = AY + CY = 6 + 7.5 = 13.5$.