The Broken Chord Theorem

Theorem

On the circumcircle of triangle $ABC$, point $P$ is the midpoint of the arc $ACB$. $PM$ is perpendicular to the longest of $AC$ or $BC$. Prove that M divides the broken line $ACB$ in half.

Proof

Let $O$ denote the circumcenter of $ΔABC$. $P'$ = symmetry of $P$ in $O$. Because $P$ is midpoint of arc $ACB$ so $PP'$ is perpendicular bisector of $AB$ and $P, C$ are on one side with respect to $AB$.

The perpendicular from $P'$ to $AC$ cuts $AC$ at $M'$ and cuts the circumcircle ($O$) again at $M''$. Two angles $CAB$ and $M''P'P$ are equal because their side lines are respectively perpendicular. Hence their subtending chords are also equal:

$CB = M''P = M'M$.

By symmetry of circumcircle, $CM = AM'$. From these:

$AM = AM' + M'M = MC + CB$

And the proof is complete.

Source: cut-the-knot