$a^2 + b^2 +c^2 +d^2 = 8r^2$ for orthodiagonal quadrilateral

Question

$\square ABCD$ orthodiagonal quadrilateral, $r$ is circumradius. Prove, that $a^2+b^2+c^2+d^2 = 8r^2$.

I proved by Pythagoras, that $a^2+c^2 =b^2+d^2$, but don`t know how to proceed from here and if my approach is good or not.

Answer 1

Wiki CyclicQuad

You are implying a cyclic quadrilateral in order to have a single $r.$

Follow equations for $D,R$ under heading circum-radius and area.

Circum-diameter $d=2r$ is expressed in terms of divided segments of orthodiagonal quadrilateral.

$$d^2= p_1^2+p_2^2+q_1^2+q_2^2,$$

$$ d= \sqrt{a^2+c^2} =\sqrt{b^2+d^2} ,\, $$

and result follows