Congratulations James Page of Hethersett High School, Norwich on your solution to this question. Again there are many possible methods of showing that the radius of the circle is equal to the side of the square.

$\angle \mathrm{EDC}={150}^{\circ }$ and $▵\mathrm{ECD}$ isosceles so $\angle \mathrm{CED}=\angle \mathrm{CDE}={15}^{\circ }$.

Now $\angle \mathrm{OED}={30}^{\circ }$ so $\angle \mathrm{OEC}=\angle \mathrm{OED}-\angle \mathrm{CED}={15}^{\circ }$ and hence the red line $\mathrm{EC}$ is a line of symmetry for the quadrilateral $\mathrm{OEDC}$ proving that the radius of the circle, $\mathrm{OC}$ is equal in length to the side of the square.