Broken Chord Theorem/Proof 3
Jump to navigation
Jump to search
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:
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:
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 |
$\blacksquare$