3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BC$ intersects circle $\omega_{1}$ again at point $P$, and line $AD$ intersects circle $\omega_{2}$ again at point $Q$. It is known that points $P, Q$, and $E$ are distinct. Find the angle $PEQ$.
Answer: $180^{\circ}$.

Solution 1. Since $\angle B C O=\angle P C O_{1}$, the isosceles triangles $B O C$ and $P O_{1} C$ are similar, from which $\angle B O C=\angle P O_{1} C$. Similarly, it can be verified that $\angle A O D=\angle Q O_{2} D$. Then the segments $O_{1} P$ and $O_{2} Q$ are parallel to the line $A B$ and, therefore, to each other. Hence, $\angle P O_{1} E=\angle Q O_{2} E$. Thus, the isosceles triangles $P O_{1} E$ and $Q O_{2} E$ are similar, from which $\angle O_{1} E P=\angle O_{2} E Q$. Therefore, the point $E$ lies on the segment $P Q$, which gives the answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-09.jpg?height=482&width=734&top_left_y=273&top_left_x=684)

Solution 2. Note a useful property of tangent circles: if a secant $U V$ passes through the point $T$ of tangency of two circles, then the inscribed angles subtended by the arcs it intercepts are equal. Indeed, since an inscribed angle is equal to the angle between the tangent and the secant, the equalities $\angle U U_{1} T=\angle U T Y=\angle X T V=\angle V V_{1} T$ hold (see the left figure below).

Let the intersection of $A D$ and $B C$ be $R$, and the second intersection of $A C$ with the circle $\omega_{1}$ be $S$. Let $\angle O O_{1} O_{2}=2 \alpha$ and $\angle O O_{2} O_{1}=2 \beta$. From the isosceles triangles $O_{1} C E$ and $O_{2} D E$, we get that

$$
\angle O_{1} E C=90^{\circ}-\alpha, \quad \angle O_{2} E D=90^{\circ}-\beta \quad \text { and } \quad \angle C E D=180^{\circ}-\angle O_{1} E C-\angle O_{2} E D=\alpha+\beta
$$

On the other hand, from the sum of the angles of triangle $O O_{1} O_{2}$, we find that $\angle C O D=180^{\circ}-2 \alpha-2 \beta$. The inscribed angle $\angle C A D$ is half the central angle $\angle C O D$. Then

$$
\angle C A D=90^{\circ}-\alpha-\beta, \quad \angle A C R=\angle A C B=90^{\circ}, \quad \text { from which } \quad \angle C R D=\alpha+\beta
$$

Thus, $\angle C R D=\angle C E D$. From the cyclic quadrilateral $P E C S$, it follows that

$$
\angle P E C=180^{\circ}-\angle P S C=180^{\circ}-\angle C A B=90^{\circ}+\angle R B A
$$

Moreover, $\angle D E Q=\angle D B A=90^{\circ}-\angle R A B$. Now we can find the angle $P E Q$:

$$
\begin{aligned}
& \angle P E Q=\angle P E C+\angle C E D-\angle D E Q=\angle P E C+\angle C R D-\angle D E Q= \\
& =\left(90^{\circ}+\angle R B A\right)+\left(180^{\circ}-\angle R A B-\angle R B A\right)-\left(90^{\circ}-\angle R A B\right)=180^{\circ}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-09.jpg?height=494&width=1322&top_left_y=1792&top_left_x=378)