In trapezoid $A B C D$, diagonal $A C$ is perpendicular to side $C D$, and diagonal $D B$ is perpendicular to side $A B$. The extensions of the lateral sides $A B$ and $D C$ intersect at point $K$, forming triangle $A K D$ with an angle of $45^{\circ}$ at vertex $K$. The area of trapezoid $A B C D$ is $P$. Find the area of triangle $A K D$.
The area of a quadrilateral is equal to half the product of its diagonals by the sine of the angle between them.

## Solution

Let $M$ be the point of intersection of the diagonals of the trapezoid. $\angle A M B = \angle A K D$ as angles with mutually perpendicular sides.

First method. $K B = B D, K C = A C$ (since triangles $D B K$ and $A C K$ are right and isosceles). Therefore,

$2 S_{B K C} = B K \cdot C K \sin \angle A K D = A C \cdot B D \sin \angle A M B = S_{A B C D} = 2 P$. Hence, $S_{A K D} = S_{B K C} + 2 S_{A B C D} = 2 P$.

Second method. Points $B$ and $C$ lie on a circle with diameter $A D$. Therefore, trapezoid $A B C D$ is isosceles. The similarity coefficient of triangles $A K D$ and $B K C$ is $\frac{A D}{B C} = \frac{A M}{M C} = \frac{A M}{M B} = \frac{1}{\cos 45^{\circ}} = \sqrt{2}$. Therefore, $S_{A K D} = 2 S_{B K C}$.

Thus, $S_{A K D} = 2 S_{A B C D} = 2 P$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-28.jpg?height=266&width=323&top_left_y=853&top_left_x=865)

Answer

$2 P$.