3. Circle $\omega$ with center $O$ touches the sides $B A$ and $B C$ of the acute angle $A B C$ at points $M$ and $N$ respectively. The line passing through point $M$ parallel to $B C$ intersects the ray $B O$ at point $K$. On the ray $M N$, a point $T$ is chosen such that $\angle M T K=\frac{1}{2} \angle A B C$. It turns out that the line $K T$ is tangent to $\omega$. Find the area of triangle $O K T$, if $B M=a$.
Answer: $\frac{a^{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-28.jpg?height=528&width=482&top_left_y=1986&top_left_x=867)

Solution. Let $L$ be the point of tangency of $K T$ with $\omega$, and $S$ be the intersection point of $M N$ with the ray $B O$. Note that triangle $M B N$ is isosceles, and $B S$ is the angle bisector of its angle $B$. Therefore, $B S$ is also the altitude and median of triangle $M B N$, so $M S = S N$ and $\angle K S M = 90^{\circ}$. The right triangles $B S N$ and $K S M$ are congruent by a leg and an angle, hence $M K = B N = a$. Moreover,

$$
\angle M K S = \angle S B C = \frac{1}{2} \angle A B C = \angle M T K
$$

Thus, triangles $M K T$ and $M S K$ are similar by two angles. Therefore, $\angle M K T = 90^{\circ}$, and triangle $M K T$ is also similar to $O N B$. Then

$$
\frac{K T}{a} = \frac{K T}{B M} = \frac{K T}{B N} = \frac{M K}{O N} = \frac{a}{O L}
$$

and the area of triangle $O T K$ is $\frac{K T \cdot O L}{2} = \frac{a^{2}}{2}$.