3. On the sides $AB$ and $CD$ of the inscribed quadrilateral $ABCD$, which is not a trapezoid, points $P$ and $R$ are marked, respectively, such that $AP: PB = CR: RD$. On the segment $PR$, a point $Q$ is chosen such that $PQ: QR = AB: CD$. Find the ratio of the areas of triangles $AQD$ and $BQC$, given that $AD = x$ and $BC = y$.
Answer: $\frac{x}{y}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-26.jpg?height=491&width=720&top_left_y=563&top_left_x=702)

Solution. Let $O$ be the point of intersection of the lines $D A$ and $C B$. For definiteness, we assume that it lies on the rays $D A$ and $C B$. Since $A B C D$ is cyclic, triangles $A O B$ and $C O D$ are similar by two angles, hence $\frac{O B}{O D}=\frac{A B}{C D}$. Moreover, by the condition

$$
\frac{P B}{A P}=\frac{R D}{C R} \Longleftrightarrow \frac{P B}{A B}=\frac{R D}{C D} \Longrightarrow \frac{P B}{R D}=\frac{A B}{C D}=\frac{O B}{O D}
$$

Finally, $\angle O D R=180^{\circ}-\angle C B P=\angle P B O$. Therefore, triangles $O B P$ and $O D R$ are similar with the ratio $\frac{A B}{C D}$. Hence, $\angle B O P=\angle D O R$ and by the condition

$$
\frac{O P}{O R}=\frac{A B}{C D}=\frac{P Q}{Q R}
$$

Thus, $O Q$ is the angle bisector of triangle $P O R$. Then

$$
\angle D O Q=\angle D O R+\angle Q O R=\angle P O Q+\angle B O P=\angle B O Q
$$

This means that point $Q$ is equidistant from the lines $A D$ and $B C$, i.e., the heights of triangles $A Q D$ and $B Q C$ dropped from point $Q$ are equal. Therefore, the ratio of the areas of these triangles is $\frac{A D}{B C}=\frac{x}{y}$.