Collinear intersection points of diagonals in a regular heptagon 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):


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*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 $0$.

*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?

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

