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 $BD$ intersects circle $\omega_{2}$ again at point $P$, and line $AC$ intersects circle $\omega_{1}$ again at point $Q$. Find the angle $PEQ$.
Answer: $180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-14.jpg?height=491&width=734&top_left_y=271&top_left_x=681)

Solution 1. Since $\angle A C O=\angle Q C O_{1}$, the isosceles triangles $A O C$ and $Q O_{1} C$ are similar, from which $\angle A O C=\angle Q O_{1} C$. Similarly, it can be verified that $\angle B O D=\angle P O_{2} D$. Then the segments $O_{1} Q$ and $O_{2} P$ are parallel to the line $A B$ and, therefore, to each other. Hence, $\angle O_{1} Q E=\angle O_{2} P E$, which is equivalent to $\angle O_{1} E Q=\angle O_{2} E P$. This means that point $E$ lies on the segment $P Q$, from which $\angle P E Q=180^{\circ}$.

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 subtending the arcs cut by it are equal. Indeed, since the 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 $R$. Let $\angle O O_{1} O_{2}=2 \alpha$ and $\angle O O_{2} O_{1}=2 \beta$. Then, 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 of the central angle $\angle C O D$. Then

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

Thus, $\angle D R C=\angle D E C$. By the proven statement,

$$
\angle P E D=\angle B A D=\angle R A B \quad \text { and } \quad \angle C E Q=\angle A B C=\angle R B A .
$$

Now we can find the angle $P E Q$ :

$$
\begin{aligned}
\angle P E Q & =\angle P E D+\angle D E C+\angle C E Q=\angle P E D+\angle D R C+\angle C E Q= \\
& =\angle P E D+180^{\circ}-\angle R A B-\angle R B A+\angle C E Q=180^{\circ} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-14.jpg?height=520&width=1314&top_left_y=1939&top_left_x=382)