3. On the side $AB$ of an acute-angled triangle $ABC$, a point $M$ is marked. A point $D$ is chosen inside the triangle. Circles $\omega_{A}$ and $\omega_{B}$ are circumscribed around triangles $AMD$ and $BMD$ respectively. The side $AC$ intersects the circle $\omega_{A}$ again at point $P$, and the side $BC$ intersects the circle $\omega_{B}$ again at point $Q$. The ray $PD$ intersects the circle $\omega_{B}$ again at point $R$, and the ray $QD$ intersects the circle $\omega_{A}$ again at point $S$. Find the ratio of the areas of triangles $ACR$ and $BCS$.
# Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-16.jpg?height=508&width=736&top_left_y=1025&top_left_x=680)

Solution. Notice that

$$
\angle P A M=180^{\circ}-\angle P D M=\angle R D M=\angle R B M
$$

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

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

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