9.5. A semicircle with diameter $A B$ and center at point $O$ is divided by points $C$ and $D$ into three parts such that point $C$ lies on the arc $A D$. Perpendiculars $D E$ and $D F$ are dropped from point $D$ to segments $O C$ and $A B$ respectively. It turns out that $D E$ is the angle bisector of triangle $A D C$, and $D O$ is the angle bisector of triangle $A D F$. Find the angle $C A D$.
Answer: $20^{\circ}$.

Solution. Triangle $A O D$ is isosceles ($O D=O A$, as radii), hence, $\angle O A D=\angle O D A$. Since $D O$ is the bisector of angle $A D F$, then $\angle O A D=\angle O D F$. Calculation of angles in the right triangle $A F D$ shows that $\angle O A D=30^{\circ}$. Let $G$ be the point of intersection of segments $A D$ and $O C$. Segment $D E$ is the height and bisector in triangle $D G C$. Therefore, triangle $D G C$ is isosceles with angle $E C D$ at the base. Triangle $O C D$ is also isosceles with angle $E C D$ at the base, hence, the angles at the vertices of these two triangles are also equal, i.e., $\angle C D G=\angle C O D$. Let $\angle C D G=\angle C O D=a$, then $\angle G C D=\angle O D C=30^{\circ}+a$. By calculating the sum of the angles in triangle $C O D$, we get that $a=40^{\circ}$. The required angle $\angle C A D$ is inscribed and subtends the same arc as the central angle $\angle D O C$, therefore, $\angle C A D=20^{\circ}$.

## Criteria.

7 points. A complete and justified solution is provided.

2 points. Only the value of $\angle O A D$ is found.