3. On the side $B C$ of triangle $A B C$ with obtuse angle $C$, a point $M$ is marked. Point $D$ is chosen such that triangle $B C D$ is acute, and points $A$ and $D$ lie on opposite sides of line $B C$. Circles $\omega_{B}$ and $\omega_{C}$ are circumscribed around triangles $B M D$ and $C M D$ respectively. Side $A B$ intersects circle $\omega_{B}$ again at point $P$, and ray $A C$ intersects circle $\omega_{C}$ again at point $Q$. Segment $P D$ intersects circle $\omega_{C}$ again at point $R$, and ray $Q D$ intersects circle $\omega_{B}$ again at point $S$. Find the ratio of the areas of triangles $A B R$ and $A C S$.
# Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-19.jpg?height=517&width=514&top_left_y=684&top_left_x=794)

Solution. Notice that

$$
\angle S B M=180^{\circ}-\angle S D M=\angle M D Q=180^{\circ}-\angle M C Q
$$

from which $S B \| A C$. Therefore, the distances from points $B$ and $S$ to line $A C$ are the same. Thus, triangles $A B C$ and $A S C$ have the same base $A C$ and equal heights dropped to $A C$. Therefore, the areas of these triangles are equal. Additionally,

$$
\angle P B M=\angle P D M=\angle R D M=\angle R C M
$$

from which $B P \| R C$. Therefore, the distances from points $R$ and $C$ to line $A B$ are the same. Thus, triangles $A B C$ and $A B R$ have the same base $A B$ and equal heights dropped to $A B$. Therefore, the areas of these triangles are also equal. As a result, we get $S_{A S C}=S_{A B C}=S_{A B R}$.