8,9 Given a parallelogram $A B C D$. A line passing through vertex $C$ intersects lines $A B$ and $A D$ at points $K$ and $L$. The areas of triangles $K B C$ and $C D L$ are $p$ and $q$. Find the area of parallelogram $A B C D$.
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.

## Solution

Consider the case when point $L$ lies on side $A D$. Let $S$ be the area of triangle $AKL$. Then the similarity coefficient of triangles $A K L$ and $B K C$ is $\frac{\sqrt{S}}{\sqrt{p}}$, and for triangles $A K L$ and $D C L$ it is $\frac{\sqrt{S}}{\sqrt{p}-\sqrt{S}}$. Therefore,

$$
S=q\left(\frac{\sqrt{S}}{\sqrt{p}-\sqrt{S}}\right)^{2}
$$

From this, we find that

$$
q=(\sqrt{p}-\sqrt{S})^{2}, \sqrt{S}=\sqrt{p}-\sqrt{q}
$$

Thus,

$$
S_{\mathrm{ABCD}}=p-S+q=p-(\sqrt{p}-\sqrt{q})^{2}+q=2 \sqrt{p q}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-26.jpg?height=352&width=879&top_left_y=744&top_left_x=592)

## Answer

## $2 \sqrt{p q}$.