Broken Chord Theorem/Proof 4
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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
Find $E$ on $BC$ such that $BD = BE$.
| \(\ds BD\) | \(=\) | \(\ds ED\) | by hypothesis | |||||||||||
| \(\ds MD\) | \(\perp\) | \(\ds BE\) | by hypothesis | |||||||||||
| \(\ds \triangle MBD\) | \(\cong\) | \(\ds \triangle MED\) | Triangle Side-Angle-Side Congruence | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds MB\) | \(=\) | \(\ds ME\) | congruence |
| \(\ds \angle MBE\) | \(=\) | \(\ds \angle MEB\) | Isosceles Triangle has Two Equal Angles | |||||||||||
| \(\ds \text {arc $AM$}\) | \(=\) | \(\ds \text {arc $MC$}\) | by hypothesis | |||||||||||
| \(\ds \angle MFC\) | \(=\) | \(\ds \angle MBC\) | Angles on Equal Arcs are Equal | |||||||||||
| \(\ds \angle AFM\) | \(=\) | \(\ds \angle MBC\) | Angles on Equal Arcs are Equal | |||||||||||
| \(\ds \angle MEB\) | \(=\) | \(\ds \angle CEF\) | Vertical Angle Theorem |
Thus we have five equal angles:
- $\angle MBE$
- $\angle MEB$
- $\angle AFM$
- $\angle MFC$
- $\angle CEF$
By Triangle with Two Equal Angles is Isosceles:
- $\triangle CEF$ is isosceles
By definition of isosceles:
- $CE = CF$
Since $\angle AFM = \angle CEF$:
- $\angle AFC + \angle ECF =$ two right angles.
| \(\ds BC\) | \(\parallel\) | \(\ds AF\) | Supplementary Interior Angles implies Parallel Lines | |||||||||||
| \(\ds AB\) | \(=\) | \(\ds CF\) | Parallel Chords Cut Equal Chords in a Circle | |||||||||||
| \(\ds AB\) | \(=\) | \(\ds CE\) | Common Notion $1$ | |||||||||||
| \(\ds AB + BD\) | \(=\) | \(\ds DE + EC\) | Common Notion $2$ | |||||||||||
| \(\ds AB + BD\) | \(=\) | \(\ds DC\) | Addition |
$\blacksquare$