Exercise 10.3.12

Proof. Let $\alpha =\mathit{\angle PRO}$. Since $\mathit{QR}=\mathit{QO}=1$ , $△\mathit{QRO}$ is isosceles, so $\mathit{\angle QOR}=\alpha$. Therefore the external angle $\mathit{\angle OQP}=2\alpha$. Since $\mathit{OQ}=\mathit{OP}=1$, $△\mathit{OPQ}$ is isosceles, so

Therefore $\mathit{\angle QOP}=\pi -4\alpha$. Since $\mathit{\angle ROQ}+\mathit{\angle QOP}+𝜃=\pi ,$ $\alpha +(\pi -4\alpha )+𝜃=\pi$, so $𝜃=3\alpha$,