Collinear intersection points of diagonals in a regular heptagon

Question

In a regular heptagon, all diagonals are drawn. Then points $A$, $B$, and $C$ in the diagram below are collinear:

(If the vertices of the pentagon are $P_1, \dots, P_7$ in clockwise order, then $A = P_1$, $B$ is the intersection of $P_2 P_5$ and $P_3 P_7$, and $C$ is the intersection of $P_2 P_4$ and $P_3 P_5$.)

Offhand, I see two approaches to trying to prove this (I verified it by direct computation in Mathematica):

1. Use the area method (Machine Proofs in Geometry style) to express the area $S_{ABC}$ in terms of areas of triangles formed by vertices of the heptagon, in hopes of showing that it's $0$.
2. Use Trig Ceva to show that $\angle P_4 AB = \angle P_4 AC$: if we draw line $AB$, then $B$ is the intersection of three chords in the circumcircle of the heptagon, forming six arcs - four of which we already know. Same for $C$.

However, both of these methods seem extremely computationally intensive. (Though I'd be happy to be proven wrong there, if someone sees a straightforward way to do either computation.)

Is there a simple proof why $A$, $B$, and $C$ are collinear, or at least a good reason why we should expect this to be true?

Answer 1

$\triangle AP_5P_7$ and $\triangle CP_2P_3$ have parallel corresponding sides, so they are homothetic about $B$; that is, $A, B, C$ are collinear.

Answer 2

In your notations, triangles $P_2P_3C$ and $P_7BP_2$ are similar because $$\angle \, CP_2P_3 = \angle \, P_4P_2P_3 = \alpha = \angle \, P_2P_7P_3 = \angle \, P_2P_7B \,\,\,\text{ and }$$ $$\angle \, P_2CP_3 = 2\, \alpha = \angle \, P_7P_2P_5 = \angle \, P_7P_2B$$ Therefore, $$\frac{CP_3}{P_2B} = \frac{P_3P_2}{BP_7}$$ However, $P_2B = P_3B $ because triangle $BP_2P_3$ is isosceles due to $$\angle \, BP_2P_3 = \angle \, P_5P_2P_3 = 2\alpha = \angle \, P_7P_3P_2 = \angle \, BP_3P_2$$ and $P_3P_2 = P_1P_7$ as edges of the regular heptagon. Thus, the latter ratio turns into $$\frac{CP_3}{P_3B} = \frac{CP_3}{P_2B} = \frac{P_3P_2}{BP_7} = \frac{P_1P_7}{BP_7}$$ or if you prefer $$\frac{CP_3}{P_1P_7} = \frac{P_3B}{BP_7}$$ Finally, observe that $P_1P_3 = P_7P_5$ so the cyclic quad $P_1P_3P_5P_7$ is an isosceles trapezoid which means that $P_1P_7$ is parallel to $P_3P_5$. Since $C$ lies on $P_3P_5$ one concludes that $CP_3$ is parallel to $P_1P_7$. Since $$\frac{CP_3}{P_1P_7} = \frac{P_3B}{BP_7}$$ and $CP_3$ is parallel to $P_1P_7$, by the Intercept Theorem, point $B$ lies on the segment $P_1C \equiv AC$.