12.171. In a right triangle $ABC$, the acute angle at vertex $A$ is $\alpha$. A line through the midpoint $D$ of the hypotenuse $AB$ intersects the leg $AC$ at point $E$. In what ratio does this line divide the area of triangle $ABC$, if $\angle DEA=\beta, AE>0.5 AC?$
Solution.

Let $S_{1}$ be the area of triangle $A D E$, $S_{2}$ be the area of quadrilateral $B C E D$, and $S$ be the area of triangle $A B C$ (Fig. 12.41).

Then

$$
\begin{aligned}
& S_{1}=\frac{1}{2} A E \cdot A D \sin \angle A=\frac{1}{4} A E \cdot A B \sin \angle A, S=\frac{1}{2} A C \cdot A B \sin \angle A \\
& \frac{S_{2}}{S_{1}}=\frac{S-S_{1}}{S_{1}}=\frac{S}{S_{1}}-1=\frac{2 A C}{A E}-1 .
\end{aligned}
$$

If $A B=c$, then from $\triangle A C B\left(\angle A C B=90^{\circ}\right): A C=c \cos \alpha$. In $\triangle A E D:$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0925.jpg?height=498&width=503&top_left_y=78&top_left_x=96)

Fig. 12.42

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0925.jpg?height=442&width=666&top_left_y=124&top_left_x=608)

Fig. 12.43

$$
\frac{A E}{\sin \angle A D E}=\frac{A D}{\sin \angle D E A} \Leftrightarrow \frac{A E}{\sin (\alpha+\beta)}=\frac{c}{2 \sin \beta} \Leftrightarrow A E=\frac{c \sin (\alpha+\beta)}{2 \sin \beta}
$$

Then $\frac{S_{2}}{S_{1}}=\frac{2 A C}{A E}-1=\frac{4 \sin \beta \cos \alpha}{\sin (\alpha+\beta)}-1=\frac{4 \sin \beta \cos \alpha-\sin (\alpha+\beta)}{\sin (\alpha+\beta)}=$

$$
=\frac{3 \sin \beta \cos \alpha-\sin \alpha \cos \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta}=\frac{3 \operatorname{tg} \beta-\operatorname{tg} \alpha}{\operatorname{tg} \alpha+\operatorname{tg} \beta}
$$

Answer: $\frac{3 \operatorname{tg} \beta-\operatorname{tg} \alpha}{\operatorname{tg} \alpha+\operatorname{tg} \beta}$.