3. On the sides $B C$ and $A D$ of the inscribed quadrilateral $A B C D$, points $K$ and $M$ are marked respectively, such that $B K: K C = A M: M D$. On the segment $K M$, a point $L$ is chosen such that $K L: L M = B C: A D$. Find the ratio of the areas of triangles $A C L$ and $B D L$, given that $A C = p$ and $B D = q$.
Answer: $\frac{p}{q}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-22.jpg?height=491&width=480&top_left_y=557&top_left_x=822)

Solution. Let $O$ be the point of intersection of $A C$ and $B D$. Since $A B C D$ is a cyclic quadrilateral, triangles $A O D$ and $B O C$ are similar by two angles, hence $\frac{O C}{O D}=\frac{B C}{A D}$. Additionally, by the given condition,

$$
\frac{K C}{B K}=\frac{M D}{A M} \Longleftrightarrow \frac{K C}{B C}=\frac{M D}{A D} \Longrightarrow \frac{K C}{M D}=\frac{B C}{A D}=\frac{O C}{O D}
$$

Finally, $\angle K C O=\angle M D O$ as inscribed angles subtending the arc $A B$. Therefore, triangles $K C O$ and $M D O$ are similar with the ratio $\frac{B C}{A D}$. Hence, $\angle K O C=\angle M O D$ and by the given condition,

$$
\frac{O K}{O M}=\frac{B C}{A D}=\frac{K L}{L M}
$$

Thus, $O L$ is the angle bisector of triangle $M O K$. Then,

$$
\angle C O L=\angle K O L-\angle K O C=\angle M O L-\angle M O D=\angle D O L
$$

This means that point $L$ is equidistant from the lines $A C$ and $B D$, so the heights of triangles $A C L$ and $B D L$ dropped from point $L$ are equal. Therefore, the ratio of the areas of these triangles is $\frac{A C}{B D}=\frac{p}{q}$.