15.5*. From the vertex $B$ of the parallelogram $A B C D$, its heights $B K$ and $B H$ are drawn. It is known that $K H=a$ and $B D=b$. Find the distance from point $B$ to the intersection point of the heights of triangle $B K H$.

 untranslated text:
15.5*. Из вершины $B$ параллелограмма $A B C D$ проведены его высоты $B K$ и $B H$. Известно, что $K H=a$ и $B D=b$. Найдите расстояние от точки $B$ до точки пересечения высот треугольника $B K H$.
15.5. Let the intersection point of the altitudes of triangle $B K H$ be denoted as $H_{1}$. Since $H H_{1} \perp B K$ and $K H_{1} \perp B H$, it follows that $H H_{1} \| A D$ and $K H_{1} \| D C$, i.e., $H_{1} H D K$ is a parallelo-

![](https://cdn.mathpix.com/cropped/2024_05_21_22b0ba5e8c5a1f49bd81g-325.jpg?height=336&width=488&top_left_y=1458&top_left_x=1076)

Fig. 15.3 gram. Therefore, under a parallel translation by the vector $\overrightarrow{H_{1} H}$, point $K$ moves to point $D$, and point $B$ moves to some point $P$ (Fig. 15.3). Since $P D \| B K$, it follows that $B P D K$ is a rectangle and $P K = B D = b$. Also, since $B H_{1} \perp K H$, it follows that $P H \perp K H$. It is clear that $P H = B H_{1}$.

In the right triangle $P K H$, the hypotenuse $K P = b$ and the leg $K H = a$ are known, so $B H_{1} = P H = \sqrt{b^{2} - a^{2}}$.