Let $T$ be a point inside a square $ABCD$. The lines $TA,TB,TC,TD$ meet the circumcircle of $ABCD$ again at $A',B',C',D'$, respectively. Prove that $A'B'\cdot C'D'=A'D'\cdot B'C'$.
Source: Slovenia National MO 2005 2nd Grade P3
Tags: geometry, square, circumcircle
Let $T$ be a point inside a square $ABCD$. The lines $TA,TB,TC,TD$ meet the circumcircle of $ABCD$ again at $A',B',C',D'$, respectively. Prove that $A'B'\cdot C'D'=A'D'\cdot B'C'$.