This problem is from the 2011/12 BMO2.
The diagonals AC and BD of a cyclic quadrilateral meet at E. The midpoints of the sides AB, BC, CD and DA are P, Q, R and S respectively. Prove that the circles EPS and EQR have the same radius.
I can't seem to find anything meaningful to work with. Would the approach be to prove that $\triangle{EPS}$ AND $\triangle{EQR}$ are congruent? I've found that PQRS is a parallelogram but I don't really know where to go from there. Please could someone provide me with a hint or starting point?
According to the diagram below, you can follow these steps:
Step $1$: $\triangle EBC$ and $\triangle EAD$ are similar. So, $\angle QEC= \angle SED.$
Step $2$: Similarly, you should be able to show that: $\angle PEA =\angle RED.$
Step $3$: $\angle QEP +\angle RES =180^{\circ} \implies \sin \angle QEP =\sin \angle RES. $
Step $4$: $|PQ|=|RS|=\frac{1}{2}|AC|.$
Step $5$: Now, by the law of sines, it follows that $\triangle EQP$ and $\triangle ERS$ have the same circumradius.