Broken Chord Theorem/Proof 1
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
Given:
- $MD \perp BC$
Find $E$ such that $BD = DE$ and draw $ME$
Then:
- $MD$ is the perpendicular bisector of $BE$.
By Triangle Side-Angle-Side Congruence:
- $\triangle MDB \cong \triangle MDE$
Let $H$ be a point such that arc $MH$ is equal to arc $BM$.
Label three angles for reference:
Let $\alpha = \angle MBC$.
Let $\beta = \angle MCB$.
Let $\gamma = \angle CMH$.
By the definition of congruence:
- $\angle MBC = \angle MEB = \alpha$
By construction:
Also by construction:
Subtracting equals:
- arc $AB = $ arc $HC$
In the words of Euclid:
- In equal circles equal circumferences are subtended by equal straight lines.
(The Elements: Book $\text{III}$: Proposition $29$)
So:
- $AB = HC$
By construction:
- arc $HC$ subtends $\gamma$
- $\angle MCH = \beta$
By construction:
- arc $MHC$ subtends $\alpha$
Equating the results:
- $\alpha = \beta + \gamma$
But $\alpha = \angle MEB$.
By External Angle of Triangle equals Sum of other Internal Angles:
- $\alpha = \angle CME + \beta$
Subtracting:
- $\angle CME = \gamma = \angle CMH$.
$MC$ is shared.
From above:
- $\angle MCH = \angle MCB = \beta$
| \(\ds \triangle MCE\) | \(\cong\) | \(\ds \triangle MCH\) | Triangle Angle-Side-Angle Congruence | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds HC\) | \(=\) | \(\ds EC\) | congruence | ||||||||||
| \(\ds AB\) | \(=\) | \(\ds HC\) | From above | |||||||||||
| \(\ds AB\) | \(=\) | \(\ds EC\) | Common Notion 1 | |||||||||||
| \(\ds AB + BD\) | \(=\) | \(\ds DE + EC\) | Common Notion 2 | |||||||||||
| \(\ds AB + BD\) | \(=\) | \(\ds CD\) | Addition |
$\blacksquare$