Let $ABCD$ be a cyclic quadrilateral. Let $C'$ be the intersection of $AC$ and $BD$. Let $B'$ and $D'$ be points on $AB$ and $AD$, respectively, such that $AB'C'D'$ be a parallelogram. Prove that $$AC.AC' = AB.AB' + AD.AD'.$$
Could somebody give me a hint? Thanks!
I just found two solutions for this problems.
The first one is to use Ptolemy's theorem and notice that $AB'C'$ and $DCB$ are similar.
The second one is to let $E, F$ be points on $AC$ such that $BCEB'$ and $DCFD'$ are cyclic quadrilaterals. Then notice that the middle points of $AC'$ and $EF$ are coincide since $B'ED'F$ is a parallelogram.