[Inscribed quadrilateral with perpendicular diagonals]

$A B C D$ is an inscribed quadrilateral with perpendicular diagonals. $O$ is the center of the circumscribed circle of quadrilateral $A B C D$. $P$ is the point of intersection of the diagonals.

Find the sum of the squares of the diagonals, given the length of the segment $O P$ and the radius of the circle $R$.
Let $M$ be the midpoint of $AC$, $N$ be the midpoint of $BD$. $AM^2 = AO^2 - OM^2$, $BN^2 = BO^2 - ON^2$, therefore $AC^2 + BD^2 = 4(R^2 - OM^2) + 4(R^2 - ON^2) = 8R^2 - 4(OM^2 + ON^2) = 8R^2 - 4OP^2$, since $OM^2 + ON^2 = OP^2$.

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