Broken Chord Theorem/Proof 3

Theorem

Let $A$ and $C$ be arbitrary points on a circle in the plane.

Let $M$ be a point on the circle with arc $AM = $ arc $MC$.

Let $B$ lie on the minor arc of $AM$.

Draw chords $AB$ and $BC$.

Find $D$ such that $MD \perp BC$.

Then:

$AB + BD = DC$

Proof

Let $E$ be a point such that $BD = DE$.

Given:

arc $AM = $ arc $MC$

By Equal Arcs of Circles Subtended by Equal Straight Lines:

$AM = MC$

By Angles on Equal Arcs are Equal:

$\angle BAM = \angle MCB$

$BM$ is shared

We have Ambiguous Case for Triangle Side-Side-Angle Congruence for these three triangles:

• $\triangle BAM$
• $\triangle MBC$
• $\triangle MEC$

Given:

$\angle MDC$ is a right angle

By External Angle of Triangle is Greater than Internal Opposite:

$\angle MEC$ is obtuse.

Since $AM = MC$ and $AC$ is the rest of the circumference:

the major arc of $AM$ is more than half the circle.

It follows that $\angle ABM$ is obtuse.

\(\ds \leadsto \ \ \)

\(\ds \triangle ABM\)

\(\cong\)

\(\ds \triangle MEC\)

\(\ds \leadsto \ \ \)

\(\ds AB\)

\(=\)

\(\ds EC\)

\(\ds AB + BD\)

\(=\)

\(\ds DE + EC = DC\)

addition

$\blacksquare$