8,9

In an isosceles trapezoid $ABCD$, the base $AD$ is equal to $a$, the base $BC$ is equal to $b$, and $AB=d$. A line through vertex $B$ bisects the diagonal $AC$ and intersects the line $AD$ at point $K$. Find the area of triangle $BDK$.

#
Quadrilateral ABCK is a parallelogram.

## Solution

Let $a>b$. Since $A B C K$ is a parallelogram, then $A K=B C=b, D K=A D-A K=a-b$.

Let $B P$ be the height of the trapezoid. Then $B P^{2}=A B^{2}-A P^{2}=d^{2}-\left(\frac{a-b}{2}\right)^{2}=\frac{1}{4}\left(4 d^{2}-(a-b)^{2}\right)$, and $S_{B D K}=\frac{1}{2} K D \cdot B P=\frac{1}{4}$

$(a-b) \sqrt{4 d^{2}-(a-b)^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-27.jpg?height=460&width=598&top_left_y=1794&top_left_x=730)

Similarly for the case $a<b$.

## Answer

$\frac{1}{4}|a-b| \sqrt{4 d^{2}-(a-b)^{2}}$.