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 $\Delta ABC$. $P^{\prime }$ = symmetry of $P$ in $O$. Because $P$ is midpoint of arc $ACB$ so $P{P}^{\prime }$ is perpendicular bisector of $AB$ and $P,C$ are on one side with respect to $AB$.

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

$CB=M^{″}P=M^{\prime }M$.

By symmetry of circumcircle, $CM=A{M}^{\prime }$. From these:

$AM=A{M}^{\prime }+M^{\prime }M=MC+CB$

And the proof is complete.

Source: cut-the-knot