Quadrilateral $ABCD$ is inscribed in a circle of radius $R$. Its diagonals are perpendicular to each other and intersect at point $P$.

Find $AP^{2}+BP^{2}+CP^{2}+DP^{2}$ and $AB^{2}+BC^{2}+CD^{2}+AD^{2}$.
Denote $\angle A D B=\alpha$ and express the sides of the quadrilateral in terms of $R$ and $\alpha$ (or draw the diameter $D E$).

## Solution

Let $\angle A D B=\alpha, \angle B D C=\beta$. Then $A B=2 R \sin \alpha, B C=2 R \sin \beta, C D=2 R \cos \alpha, A D=2 R \cos \beta$.

Therefore, $A P^{2}+B P^{2}+C P^{2}+D P^{2}=A B^{2}+D C^{2}=4 R^{2} \sin ^{2} \alpha+4 R^{2} \cos ^{2} \alpha=4 R^{2}, A B^{2}+B C^{2}+C D^{2}+A D^{2}=4 R^{2}+$ $4 R^{2}=8 R^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-31.jpg?height=380&width=395&top_left_y=456&top_left_x=840)

## Answer

$4 R^{2}, 8 R^{2}$.