3. The diagonals of the inscribed quadrilateral $ABCD$ intersect at point $O$. Inside triangle $AOB$, a point $K$ is chosen such that line $KO$ is the bisector of angle $CK$. Ray $DK$ intersects the circumcircle of triangle $COK$ again at point $L$, and ray $CK$ intersects the circumcircle of triangle $DOK$ again at point $M$. Find the ratio of the areas of triangles $ALO$ and $BMO$.
Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-02.jpg?height=497&width=1148&top_left_y=1008&top_left_x=474)

Solution. Let $r_{1}$ and $r_{2}$ be the radii of the circumcircles of triangles $C O K$ and $D O K$, respectively. Note that

$$
\angle L K O=180^{\circ}-\angle D K O=180^{\circ}-\angle C K O=\angle M K O
$$

from which $\frac{L O}{M O}=\frac{r_{1}}{r_{2}}$. Moreover, the inscribed quadrilateral $A B C D$ implies that triangles $A O D$ and $B O C$ are similar by two angles. Then

$$
\frac{A O}{B O}=\frac{O D}{O C}=\frac{r_{2}}{r_{1}}
$$

since the chords $O D$ and $O C$ correspond to equal inscribed angles. Therefore,

$$
\frac{L O}{M O} \cdot \frac{A O}{B O}=1 \Longleftrightarrow L O \cdot A O=M O \cdot B O
$$

Since $\angle K M O=\angle K D O$ and $\angle K L O=\angle K C O$, triangles $M C O$ and $D L O$ are similar, from which

$$
\begin{gathered}
\angle A O L=\angle A O M+\angle L O M=180^{\circ}-\angle C O M+\angle L O M= \\
\quad=180^{\circ}-\angle D O L+\angle L O M=\angle B O L+\angle L O M=\angle B O M
\end{gathered}
$$

Thus,

$$
S_{A O L}=\frac{1}{2} \cdot L O \cdot A O \cdot \sin \angle A O L=\frac{1}{2} \cdot M O \cdot B O \cdot \sin \angle B O M=S_{B O M}
$$