8.5. In an isosceles triangle \(ABC\), the angle \(A\) at the base is \(75^\circ\). The bisector of angle \(A\) intersects side \(BC\) at point \(K\). Find the distance from point \(K\) to the base \(AC\), if \(BK=10\).
Answer: 5.

Solution. In triangle $ABC$, $\angle B = (180^0 - 75^0 - 75^0) = 30^0$. Drop perpendiculars from point $K$: $KH$ to side $AB$, and $KN$ to side $AC$. In the right triangle $HBK$, the hypotenuse $BK$ is 10. Angle $B$ is 30 degrees, and the side opposite to it is half the hypotenuse, so $HK = 5$. Triangles $AHK$ and $ANK$ are equal by hypotenuse and acute angle, so $KN = KH = 5$.

## Criteria.

1 point. A perpendicular is dropped from point $K$ to side $AB$.

2 points. The length of the perpendicular dropped from point $K$ to side $AB$ is found. Comment. Points for criteria are summed.