In the right triangle $ABC$, the foot of the altitude from $C$ to the hypotenuse $AB$ is $D$. The angle bisectors of $\angle ACD$ and $\angle BCD$ intersect the hypotenuse $AB$ at points $E$ and $F$, respectively. Determine the ratio of the radius of the incircle of triangle $ABC$ to the radius of the circumcircle of triangle $CEF$.
Solution. First, we show that the centers of the two circles coincide. Let $2 \alpha$ be the angle at $A$ in triangle $ABC$, and let $O$ be the intersection of the angle bisectors of the triangle. Since the acute angles of right-angled triangles are equal, we have $\angle CAD = \angle BCD = 2 \alpha$ (Figure 1), so

$$
\angle FCA = \angle FCD + \angle DCA = \alpha + (90^\circ - 2\alpha) = 90^\circ - \alpha
$$

In triangle $AFC$, the sum of the angles is $180^\circ$, so

$$
\angle AFC = 180^\circ - (\angle FCA + \angle CAF) = 90^\circ - \alpha.
$$

Therefore, triangle $AFC$ is isosceles, which implies that the angle bisector of its vertex angle is perpendicular to the base. Thus, the perpendicular bisector of segment $CF$ is the line $AO$. Similarly, we find that the perpendicular bisector of segment $CE$ is the line $BO$. Since the intersection of two perpendicular bisectors of sides determines the center of the circumcircle of triangle $CEF$, it coincides with $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_fe2d9d1e0f9121ef4a11g-1.jpg?height=388&width=596&top_left_y=711&top_left_x=754)

Let $P$ and $Q$ be the points of tangency of the incircle of triangle $ABC$ on the legs (Figure 2). Since the lengths of the two tangents drawn from any external point to a circle are equal, we have $CP = CQ$. Clearly, $OP = OQ$ also holds, so quadrilateral $CPOQ$ is a kite. The tangent to a circle is perpendicular to the radius at the point of tangency, and $\angle PCQ = 90^\circ$, so the kite has three right angles, making it a rectangle. If a kite is a rectangle, then it is a square.

![](https://cdn.mathpix.com/cropped/2024_05_02_fe2d9d1e0f9121ef4a11g-1.jpg?height=393&width=593&top_left_y=1332&top_left_x=755)

The desired ratio of the radii of the two circles is therefore the same as the ratio of the side $OP$ to the diagonal $OC$ of the square $CPOQ$, which is $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.