KöMaL 2010 B4289

The diagonals of a trapezium $A_1 A_2 A_3 A_4$ are $A_1 A_3$=$e$ and $A_2 A_4$=$f$. Let $r_i$ denote the radius of the circumscribed circle of triangle $A_j A_k A_l$, where {1,2,3,4}={$i$,$j$,$k$,$l$}. Show that $\frac{r_2+r_4}{e}=\frac{r_1+r_3}{f}$ .