Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0973a8d23c1bf92cb27dg-24.jpg?height=378&width=422&top_left_y=1434&top_left_x=512)
Answer: 21.

Solution. Since $AC$ is the diameter of the circle, point $O$ is the midpoint of $AC$, and $\angle AKC = 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0973a8d23c1bf92cb27dg-25.jpg?height=371&width=407&top_left_y=196&top_left_x=517)

Then

$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA + \angle CAK = \angle BKA = 90^{\circ}.
$$

The area of the right triangle $ABC$ can be found as $\frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} \cdot 6 \cdot 14 = 42$. Since the median $BO$ divides its area in half, the area of triangle $BOC$ is $\frac{42}{2} = 21$.