3. On the sides $AB$ and $AC$ of an acute-angled triangle $ABC$, points $K$ and $L$ are marked such that the quadrilateral $B K L C$ is cyclic. Inside this quadrilateral, a point $M$ is chosen so that the line $AM$ is the bisector of angle $BMC$. The ray $BM$ intersects the circumcircle of triangle $AMC$ again at point $P$, and the ray $CM$ intersects the circumcircle of triangle $AMB$ again at point $Q$. Find the ratio of the areas of triangles $ALP$ and $AKQ$.
Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-05.jpg?height=508&width=785&top_left_y=1025&top_left_x=658)

Solution. Let $r_{1}$ and $r_{2}$ be the radii of the circumcircles of triangles $A M B$ and $A M C$, respectively. Since $A M$ is the bisector of angle $B M C$, the following equalities hold:

$$
\angle A M B = \angle A M C \quad \text{and} \quad \angle A M P = \angle A M Q
$$

from which $\frac{A B}{A C} = \frac{r_{1}}{r_{2}} = \frac{A Q}{A P}$. Moreover, from the inscribed quadrilateral $B K L C$, it follows that

$$
\angle A L K = 180^{\circ} - \angle K L C = \angle A B C
$$

Thus, triangles $A L K$ and $A B C$ are similar by two angles. Therefore,

$$
\frac{A L}{A K} = \frac{A B}{A C} = \frac{A Q}{A P} \Longrightarrow A L \cdot A P = A K \cdot A Q
$$

From the equalities $\angle A B M = \angle A Q M$ and $\angle A P M = \angle A C M$, it follows that $\angle B A P = \angle C A Q$, from which

$$
\angle L A P = \angle C A P = \angle B A P - \angle B A C = \angle C A Q - \angle B A C = \angle B A Q = \angle K A Q
$$

Thus,

$$
S_{A L P} = \frac{1}{2} \cdot A L \cdot A P \cdot \sin \angle L A P = \frac{1}{2} \cdot A K \cdot A Q \cdot \sin \angle K A Q = S_{A K Q}
$$