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

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The distance from the vertex of a triangle to the point of intersection of the altitudes is twice the distance from the center of the circumscribed circle to the opposite side.

## Solution

## First Method.

Since the diagonal $B D$ of the given parallelogram is seen from points $K$ and $H$ at a right angle, points $K$ and $H$ lie on a circle with diameter $B D$. Let $P$ be the projection of the center $O$ of this circle ( $O$ is the midpoint of $B D$ ) onto the side $K H$ of triangle $B K H$, and $M$ be the point of intersection of the altitudes of triangle $B K H$ (Fig.1). Then

$$
O P=\sqrt{O H^{2}-P H^{2}}=\sqrt{\left(\frac{b}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}}=\frac{\sqrt{b^{2}-a^{2}}}{2}
$$

Therefore, $B M=2 O P=\sqrt{b^{2}-a^{2}}$.

## Second Method.

Let $F$ be the projection of point $D$ onto the line $B C$, and $M$ be the point of intersection of the altitudes of triangle $B K H$ (Fig.2). When the vector $\overrightarrow{K D}$ is translated parallel, point $B$ moves to point $F$, and point $M$ moves to point $H$ (since $K M H D$ is a parallelogram). Therefore, triangle $F H D$ is equal to triangle $B M K$.

In the right triangle $K H F$, the leg $K H=a$ and the hypotenuse $K F=B D=b(K B F D$ is a rectangle). Therefore,

$$
B M=F H=\sqrt{K F^{2}-K H^{2}}=\sqrt{b^{2}-a^{2}} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6c7d8d8bac098e64f10bg-43.jpg?height=345&width=521&top_left_y=-1&top_left_x=454)

Fig. 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6c7d8d8bac098e64f10bg-43.jpg?height=301&width=519&top_left_y=18&top_left_x=1090)

Fig. 2.

## Answer

$\sqrt{b^{2}-a^{2}}$.