## Inscribed Quadrilateral with Perpendicular Diagonals Pythagorean Theorem (direct and inverse) Ratio of Areas of Similar Triangles ] Area of a Quadrilateral ] Complex ]  [ Ratio of Areas of Triangles with a Common Base or Height]

Diagonals $A C$ and $B D$ of an inscribed quadrilateral intersect at point $Q$ at a right angle. Lines $A B$ and $C D$ intersect at point $P$. It is known that $B C=5, A D=10, B Q=3$. Find $A P$.

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From the right triangle $B Q C$ we find that $C Q=4$.

Right triangles $A Q D$ and $B Q C$ are similar with a ratio of $A D / B C=2$, so $A Q=6, D Q=8$. By the Pythagorean theorem,

$A B^{2}=B Q^{2}+A Q^{2}=45$.

$S_{A B C D}=1 / 2 A C \cdot B D=55, S_{A B C}=1 / 2 A C \cdot B Q=15$.

Triangle $B P C$ is similar to triangle $D P A$ by two angles with a similarity ratio of ${ }^{B C} / A D=1 / 2$, so $S_{A B C D}=3 / 4 S_{A P D}=3 S_{B P C}$, from which

$A P: A B=\left(S_{A B C}+S_{A P D}\right): S_{A B C}=(45+55): 45=20: 9$, that is, $A P=\frac{20}{9} \cdot 3 \sqrt{5}=\frac{20 \sqrt{5}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-06.jpg?height=557&width=391&top_left_y=1728&top_left_x=845)

## Answer

$\frac{20 \sqrt{5}}{3}$.