3. For a convex quadrilateral $A B C D$, it is known that $A B=B C=C A=$ $C D, \angle A C D=10^{\circ}$. A circle $\omega$ is circumscribed around triangle $B C D$ with center $O$. Line $D A$ intersects circle $\omega$ at points $D$ and $E$. Find the measure of angle $E O A$, express your answer in degrees.
Answer: 65.

![](https://cdn.mathpix.com/cropped/2024_05_06_2b931cd4ed69e2b0babdg-2.jpg?height=395&width=408&top_left_y=1290&top_left_x=529)

Fig. 2: to the solution of problem 3.

Fig. 2. Since triangle $A D C$ is isosceles, and we know the angle at its vertex, the angles at its base are $\angle D A C = \angle C D A = 85^{\circ}$. Therefore, $\angle B A E = 180^{\circ} - \angle D A C - \angle C A B = 35^{\circ}$. Angle $E B C$ complements angle $E D C$ to $180^{\circ}$ and is thus $95^{\circ}$, from which $\angle E B A = 35^{\circ}$. Consequently, triangle $A E B$ is isosceles, and line $C E$ is the perpendicular bisector of $A B$. But then $C B$ is the angle bisector of $\angle B C A$, so $\angle B C E = 30^{\circ}$, and the corresponding central angle $\angle B O E = 60^{\circ}$.

On the other hand, $\angle C O E = 2 \angle C D E = 170^{\circ}$, from which $\angle C O B = 110^{\circ}$.

Finally, from the symmetry of triangle $A B C$ with respect to $A O$, it is easy to see that the desired angle $\angle E O A$ complements to $180^{\circ}$ the sum of angle $\angle B O E$ and half of angle $\angle C O B$, i.e., $\angle E O A = 180^{\circ} - \angle B O E - \frac{1}{2} \angle C O B = 65^{\circ}$.