1. A line is drawn through the vertex $C$ of parallelogram $A B C D$ which intersects the rays $A B$ and $A D$ at points $K$ and $L$ respectively. Express the area $s$ of parallelogram $A B C D$ in terms of the areas $p$ and $q$ of triangles $BKC$ and $D C L$.
Solution. The corresponding angles of triangles BKC and $D C L$ are equal, so $\triangle B K C \sim \triangle D C L$. Therefore,

$$
h_{1}: h_{2}=\overline{B K}: \overline{D C}
$$

i.e., $h_{1} \cdot \overline{D C}=h_{2} \cdot \overline{B K}$. If we multiply the last equality by $h_{1} \cdot \overline{D C}$ and use the fact that

$$
h_{1} \cdot \overline{D C}=s, h_{1} \cdot \overline{B K}=2 p, h_{2} \cdot \overline{D C}=2 q
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e63ee82c9277009fa2feg-1.jpg?height=473&width=485&top_left_y=566&top_left_x=995)
we get

$$
s^{2}=\left(h_{1} \cdot \overline{D C}\right)^{2}=\left(h_{1} \cdot \overline{D C}\right) \cdot\left(h_{2} \cdot \overline{B K}\right)=\left(h_{1} \cdot \overline{B K}\right) \cdot\left(h_{2} \cdot \overline{D C}\right)=4 p q
$$

i.e., $s=2 \sqrt{p q}$.