## Task 5 - 090935

The area of triangle $\triangle A B C$ is divided into two equal areas by a line parallel to side $A B$.

Determine the ratio in which the height of the triangle corresponding to side $A B$ is divided by the parallel line!
Let $P$ and $Q$ be the intersection points of the parallel lines with the sides $A C$ and $B C$, respectively. Furthermore, let $h_{A B}$ and $h_{P Q}$ be the lengths of the heights from point $C$ to $A B$ and the parallel line $P Q$, respectively.

According to the problem statement, the area $A_{P Q C}$ of triangle $\triangle P Q C$ is exactly half the area $A_{A B C}$ of triangle $\triangle A B C$.

Thus,

$$
\frac{1}{2}|P Q| \cdot h_{P Q}=A_{P Q C}=\frac{1}{2} A_{A B C}=\frac{1}{4} \cdot|A B| \cdot h_{A B}
$$

or

$$
2=\frac{|A B|}{|P Q|} \cdot \frac{h_{A B}}{h_{P Q}}
$$

By the intercept theorem, $\frac{|A B|}{|P Q|}=\frac{h_{A B}}{h_{P Q}}$, so $h_{A B}=\sqrt{2} \cdot h_{P Q}$. Therefore, the height $h_{A B}$ is divided by the parallel line $P Q$ to $A B$ in the ratio $1:(\sqrt{2}-1)$.