A convex quadrilateral's diagonals form a $45^{\circ}$ angle. Drop a perpendicular from each vertex of the quadrilateral to the line of the diagonal connecting the two adjacent vertices. How does the area of the quadrilateral formed by the feet of the perpendiculars compare to the area of the original quadrilateral?
Solution. Let the vertices of the quadrilateral be $A, B, C, D$, the feet of the perpendiculars be $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$, and the intersection of the diagonals be $K$. In the right triangle $A A^{\prime} K$,

$$
\cos 45^{\circ}=\frac{K A^{\prime}}{K A}, \quad \text { from which } \quad K A^{\prime}=\frac{\sqrt{2}}{2} K A
$$

Similarly,

$$
K B^{\prime}=\frac{\sqrt{2}}{2} K B, \quad K C^{\prime}=\frac{\sqrt{2}}{2} K C
$$

and

$$
K D^{\prime}=\frac{\sqrt{2}}{2} K D
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_346873209cca15f32ed6g-1.jpg?height=688&width=553&top_left_y=667&top_left_x=775)

$\triangle D^{\prime} K A^{\prime} \sim \triangle D K A$, since the ratio of two sides and the included angle are the same. The similarity ratio is $\frac{\sqrt{2}}{2}$, and the ratio of the areas of the two triangles is the square of this, which is $\frac{1}{2}$.

Similarly, $\triangle A^{\prime} K B^{\prime} \sim \triangle A K B$, $\triangle B^{\prime} K C^{\prime} \sim \triangle B K C$, and $\triangle C^{\prime} K D^{\prime} \sim \triangle C K D$, with the similarity ratio being $\frac{\sqrt{2}}{2}$ and the area ratio being $\frac{1}{2}$ in each case.

Since the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the sum of the areas of the four small triangles, and the area of $A B C D$ is the sum of the areas of the four large triangles, the ratio of the areas of the quadrilaterals is also $\frac{1}{2}$.