In quadrilateral $A B C D$, the acute angle between the diagonals is $\alpha$. Through each vertex, a line is drawn perpendicular to the diagonal that does not contain this vertex. Find the ratio of the area of the quadrilateral bounded by these lines to the area of quadrilateral $A B C D$.
If $d_{1}$ and $d_{2}$ are the diagonals of a quadrilateral, then the heights of the resulting parallelogram are $d_{1} \cos \alpha$ and $d_{2} \cos \boldsymbol{\alpha}$.

## Solution

Let $d_{1}$ and $d_{2}$ be the diagonals of the quadrilateral $A B C D$. The new quadrilateral is a parallelogram. Its heights are equal to the projections of the diagonals $A C$ and $B D$ onto each other, i.e., $d_{1} \cos \alpha$ and $d_{2} \cos \alpha$. Denote the area of this parallelogram by $S^{\prime}$.

Since the area of a parallelogram is equal to the product of its two heights divided by the sine of the angle between them (which is equal to the angle between the sides), we have

$$
S^{\prime}=\frac{d_{1} \cos \alpha \cdot d_{2} \cos \alpha}{\sin \alpha}=\frac{d_{1} d_{2} \cos ^{2} \alpha}{\sin \alpha}
$$

Therefore,

$$
\frac{S^{\prime}}{S_{A B C D}}=\frac{\frac{d_{1} d_{2} \cos ^{2} \alpha}{\sin \alpha}}{\frac{1}{2} d_{1} d_{2} \sin \alpha}=2 \cot^{2} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-12.jpg?height=415&width=648&top_left_y=730&top_left_x=705)

## Answer

$2 \cot^{2} \alpha$