Task 14. Find the ratio of the legs $CB$ and $CA$ of the right triangle $ACB$, if it is known that one half of the hypotenuse (from the midpoint of the hypotenuse to the vertex) is seen from the center $X$ of the inscribed circle at a right angle.
Solution. The angle $A X B$ is equal to $135^{\circ}$ (Fig. 13). Therefore, point $X$ is the intersection of the circumcircles of triangle $A X B$ and the right triangle $A X O(\overline{A O}=\overline{O B})$. After this, the construction of a triangle similar to triangle $A C B$ is clear.

We construct an arbitrary segment $A B$. We divide it at point $O$ into two equal parts. We construct a circle $\omega$ with radius $O A$ and center at point $O$. We construct an arc $\omega_{1}$ of a circle that is subtended by the chord $A B$ and in which an angle of $135^{\circ}$ can be inscribed. We construct a semicircle $\omega^{1}$ with diameter $A O$. Point $X$ is the intersection of figures $\omega$ and $\omega^{\prime}$. We draw a line through point $O$ perpendicular to line $A B$, and on this line, we mark a segment $O M$,

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-064.jpg?height=597&width=680&top_left_y=2003&top_left_x=1039)
equal to $A O$ (point $M$ does not lie on the semicircle containing figure $\omega_{1}$). The third vertex $C$ of the right triangle $A C B$ is the intersection of line $M X$ with circle $\omega_{1}$.

Let $A(0 ; 0), O(1 ; 0), B(2 ; 0)$, and $M(1 ;-1)$. Then the equation of the circumcircle of the right triangle $O X A$ is:

$$
(x-0.5)^{2}+y^{2}=0.25
$$

We find the radius $R$ of the circumcircle of triangle $A X B$:

$$
R=A B: 2 \sin 135^{\circ}=\sqrt{2}
$$

After this, it is clear that the center of this circle is point $M(1 ;-1)$, and its equation is:

$$
(x-1)^{2}+(y+1)^{2}=2
$$

Solving the system of equations (1) and (2), we get $X(0.8 ; 0.4)$. This means that $\operatorname{tg} \alpha=\operatorname{tg} \angle X A B=0.5$ and $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{4}{3}$.

From this, $C B: C A=4: 3$.