parmenides51 wrote: A regular heptagon $ABCDEFG$ is given. The lines $AB$ and $CE$ intersect at $P$. Find the measure of the angle $\angle PDG$. Since it is regular we have its angles are $\frac{5\pi}{7}$. Now, $DE=DC \Longrightarrow \angle DCE=\frac{\pi-\frac{5\pi}{7}}{2}=\frac{\pi}{7}$. $\angle BCP = \pi - \angle BCE = \pi - (\frac{5\pi}{7}-\angle DCE) = \frac{3\pi}{7}$. $\angle PBC = \pi-\frac{5\pi}{7}=\frac{2\pi}{7}$. $\angle BPC = \pi - \angle PCB - \angle PBC = \frac{2\pi}{7} \Longrightarrow BC = CP \Longrightarrow C$ is the circumcenter of $\triangle BPD \Longrightarrow \angle PDB = \frac{\angle PCB}{2} = \frac{3\pi}{14}$. $ABCDEFG$ is cyclic $\Longrightarrow \angle BDG = \pi - \angle BAG = \frac{2\pi}{7}$. $\angle PDG = \angle PDB + \angle BDG = \frac{3\pi}{14} + \frac{2\pi}{7} = \boxed{\frac{\pi}{2}}$.