সিমসনের উপপাদ্যের বর্ধন

The Simson Line

Given a triangle ABC and a point P on its circumcircle, the three closest points to P on lines AB, AC, and BC are collinear.

Proof: To prove this, we angle chase. It suffices to show that: $$\angle AEF = \angle CED$$

Since: $$\angle PEA = \angle PFA = 90^\circ$$ We get that quadrilateral $PAFE$ is cyclic. Similarly, we have that quadrilaterals $PDBF$ and $PDCE$ are cyclic. Because of the cyclic quadrilaterals $PDBF$ and $PABC$, we have: $$\angle DPF = \angle ABC = \angle APC$$ We can subtract $\angle FPC$ from both sides to get: $$\angle CPD = \angle APF$$ Since $PDCE$ is cyclic, we have: $$\angle CPD = \angle CED$$ Since $PAFE$ is cyclic, we have: $$\angle APF = \angle AEF$$ So, using the transitive property, we have: $$\angle AEF = \angle CED$$ And we are done.