This is problem 365 from Kiselev's Planimetry book. I have to show that two inscribed quadrilaterals with perpendicular diagonals are similar iff they have respectively congruent angles. Here is my proof which I believe may have a mistake:
Let $FCEB$ and $GHIJ$ be the respective quadrilaterals. If they are similar then by definition they have congruent angles, so lets jump to the second part: assume they have respectively congruent angles. Since they both have perpendicular diagionals I can pick up $GHIJ$ and superimpose it so that the intersection of the diagonals coincide and the diagonals lie on the same lines as shown in the diagram. Then if we translate $GHIJ$ by the vector $GF$, the image of $JH$ would become parallel to $BC$ and thus $\frac{GJ}{FB} = \frac{GH}{FC}= \frac{JH}{BC}$. Finally, if we translate $GHIJ$ by the vector $IE$ we have $\frac{IJ}{EB} = \frac{IH}{EC} = \frac{JH}{BC}$ and we see that these equaltions of ratios are connected by $\frac{JH}{BC}$ from where it follows that the quadrilaterals are similar.
Here is a diagram showing what happens after $GHIJ$ is translated.$FH'J'$ and $FCB$ are similar from where the ratio equalities follow.
I think the proof has a mistake as we didnt use the fact that the quadrilaterals are inscribed. So I am asking for both proof verification and alternative proofs.