Figure 2: Problem 10

As usual, we denote the angles of the triangle at $A$, $B$ and $C$ with $\alpha$, $\beta$ and $\gamma$.
It is sufficient to show that the center $I_c$ of the excircle touching the line $AB$ lies on the two circles. To do this, we look at the respective inscribed angles.
Since the triangle $AYB$ is isosceles, the following holds:
$$\angle CYA = \angle BYA = 90^\circ - \angle YBA/2 = 90^\circ - 90^\circ + \beta/2 = \beta/2.$$

But it is also true that
$$\angle CI_c A = 180^\{\circ\} - (180^\{\circ\} - \alpha)/2 - \alpha - \gamma/2 = 90^\{\circ\} - \alpha/2 - \gamma/2 = \beta/2.$$
So $I_c$ lies on the circumcircle of $ACY$ by the inverse of the inscribed angle theorem. In the same way, one also obtains that $I_c$ lies on the circumcircle of $BCX$. So $I_c$ is the second point of intersection, which therefore lies on the angle bisector through $C$ as required.