2004 USAMO 第 6 题

A circle $\omega$ is inscribed in a quadrilateral $ABCD$. Let $I$ be the center of $\omega$. Suppose that

$$(AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2.$$

Prove that $ABCD$ is an isosceles trapezoid.

圆 $\omega$ 内接于四边形 $ABCD$。设$I$为$\omega$的中心。假设

$$(AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2。$$

证明$ABCD$是等腰梯形。