Let $ABCD$ be a cyclic quadrilateral and triangles $ACD, BCD$ are acute. Suppose that the lines $AB$ and $CD$ meet at $S$. Denote by $E$ the intersection of $AC, BD$. The circles $(ADE)$ and $(BC E)$ meet again at $F$. a) Prove that $SF \perp EF.$ b) The point $G$ is taken out side of the quadrilateral $ABCD$ such that triangle $GAB$ and $FDC$ are similar. Prove that $GA+ FB = GB + FA$