## [Central symmetry helps solve the task]

$A B C D$ - is the given parallelogram. Through the point of intersection of its diagonals, a line perpendicular to $B C$ is drawn, which intersects $B C$ at point $E$, and the extension of $A B$ - at point $F$. Find $B E$, if $A B=a, B C=b$ and $B F=c$.

#
If $K$ is the intersection point of lines $E F$ and $A D$, then triangles $F B E$ and $F A K$ are similar.

## Solution

Let $O$ be the intersection point of the diagonals, $K$ be the intersection point of lines $E O$ and $A D$, and $x$ be the desired segment. Segments $K D$ and $B E$ are symmetric with respect to point $O$, so $K D = B E = x$, and $A K = A D - K D = b - x$. From the similarity of triangles $F B E$ and $F A K$, it follows that $B E : A K = B F : A F$, or $\frac{x}{b-x} = \frac{c}{a+c}$. Therefore, $x = \frac{b c}{a + 2 c}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-34.jpg?height=346&width=548&top_left_y=730&top_left_x=755)

## Answer

$\frac{b c}{a + 2 c}$