I've been working on part 2 of this problem for a while (I managed to finish part 1 (prove that $\angle ARP \simeq \angle QRB$ )) and I just can't seem to find a way to start calculating the ratios of $\triangle QRB$ and $\triangle AQP$. I would really appreciate any help on the problem - I'm really stuck!
Thanks.
Prove that $\triangle AEC$ is similar to $\triangle BDE$.
These triangles are similar, since they share the angle $AEC$ and $\angle CAE=\angle EBD=45^\circ$, since
\begin{align} \angle DBA&=90^\circ ,\quad \angle ABC=\tfrac12\angle AOC=45^\circ ,\\ \angle EBD&=180^\circ-\angle DBA-\angle ABC=45^\circ . \end{align}
It follows that $\angle ECA=\angle BDE$ and hence, triangles $ABC$ and $BDE$ are also similar.
When $|DE|=|OD|=R$ we have $R_{ABC}=R$ as a radius of the circumscribed circle of $\triangle ABC$, and by the sine rule the circumradius of $\triangle BDE$ is found as
\begin{align} R_{BDE}&=\frac{|DE|}{2\sin\angle EBD} =\frac{R}{2\sin 45^\circ} =\tfrac{\sqrt2}2\,R ,\\ \end{align}
Hence, the ratio of the areas of $\triangle ABC$ and $\triangle BDE$ is \begin{align} \frac{S_{ABC}}{S_{BDE}} &= \frac{R_{ABC}^2}{R_{BDE}^2} =\frac{R^2}{(\tfrac{\sqrt2}2\,R)^2} =2 . \end{align}