Source: Czech-Polish-Slovak Match Junior 2021, individual p5 CPSJ
Tags: angles, hexagon, geometry
A regular heptagon $ABCDEFG$ is given. The lines $AB$ and $CE$ intersect at $ P$. Find the measure of the angle $\angle PDG$.
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 $\Longrightarrow \angle BDG = \pi - \angle BAG = \frac{2\pi}{7}$.
$\angle PDG = \angle PDB + \angle BDG = \frac{3\pi}{14} + \frac{2\pi}{7} = \boxed{\frac{\pi}{2}}$.