If the diagonal AC and BD of the quadrilateral ABCD inscribed by the circle O intersect at the K point vertically, let the radius of the circle O be R, and prove that AK 2+BK 2+CK 2+DK 2 is constant.

Let the distance from the center o to AC be a.

The distance from the center o to BD is B.

rule

AK=√(R^2-a^2 ) +b

CK=√(R^2-a^2 )-b

BK=√(R^2-b^2 )+a

Dk = √ (r 2-b 2)-A.

AK？ +BK？ +CK？ +DK？ = R？ -a？ + b？ +2b√(R^2-a^2 )+ R？ -a？ + b？ -2b√(R^2-a^2 )+ R？ -B? + a？ +2a√(R^2-b^2 )+ R？ -B? + a？ -2a√(R^2-b^2 )=4R？

AB^2+BC^2+CD^2+DA^2=2(AK？ +BK？ +CK？ +DK？ )=8R？