Broken Chord Theorem
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 1
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 |
$\blacksquare$
Proof 2
Let point $E$ be such that $BD = DE$.
Extend $BC$ to $G$ such that $GD = DC$.
| \(\ds \triangle MDB\) | \(\cong\) | \(\ds \triangle MDE\) | Triangle Side-Angle-Side Congruence | |||||||||||
| \(\ds \triangle MDG\) | \(\cong\) | \(\ds \triangle MDC\) | Triangle Side-Angle-Side Congruence | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds MG\) | \(=\) | \(\ds MC\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \angle MGC\) | \(=\) | \(\ds \angle MCG\) |
Given:
| \(\ds AM\) | \(=\) | \(\ds MC\) | Equal Arcs of Circles Subtended by Equal Straight Lines | |||||||||||
| \(\ds AM\) | \(=\) | \(\ds MG\) | Common Notion 1 |
By the definition of isosceles triangles:
- $\triangle MGA$ is isosceles.
| \(\ds \angle MGA\) | \(=\) | \(\ds \angle MAG\) | Isosceles Triangle has Two Equal Angles | |||||||||||
| \(\ds \angle MCG\) | \(=\) | \(\ds \angle MAB\) | Angles on Equal Arcs are Equal | |||||||||||
| \(\ds \angle MGC\) | \(=\) | \(\ds \angle MAB\) | Common Notion 1 | |||||||||||
| \(\ds \angle BGA\) | \(=\) | \(\ds \angle BAG\) | Common Notion 3 |
By Triangle with Two Equal Angles is Isosceles:
- $\triangle BAG$ is isosceles.
| \(\ds AB\) | \(=\) | \(\ds GB\) | isosceles triangles | |||||||||||
| \(\ds GD\) | \(=\) | \(\ds DC\) | By construction | |||||||||||
| \(\ds GB + BD\) | \(=\) | \(\ds DE + EC\) | Common Notion 2 | |||||||||||
| \(\ds AB + BD = DE + EC\) | \(=\) | \(\ds DE + EC\) | Common Notion 1 | |||||||||||
| \(\ds AB + BD\) | \(=\) | \(\ds DC\) | Addition |
$\blacksquare$
Proof 3
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 |
$\blacksquare$
Proof 4
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 |
$\blacksquare$
Proof 5
Given $MD \perp BC$
Draw $MN \parallel BC$ to meet the circle at $N$.
Draw $NE \parallel MD$.
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
- $MNED$ is a parallelogram
By Parallelogram with One Right Angle is Rectangle:
- $MNED$ is a rectangle.
| \(\ds DE\) | \(=\) | \(\ds MN\) | Definition of Rectangle | |||||||||||
| \(\ds MD\) | \(=\) | \(\ds NE\) | Definition of Rectangle |
By Parallelism implies Equal Corresponding Angles:
- $\angle MDB = \angle NEC$
and both are right angles.
| \(\ds BM\) | \(=\) | \(\ds NC\) | Parallel Chords Cut Equal Chords in a Circle | |||||||||||
| \(\ds \triangle MDB\) | \(\cong\) | \(\ds \triangle NEC\) | Triangle Side-Side-Side Congruence | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds BD\) | \(=\) | \(\ds EC\) | |||||||||||
| \(\ds \text {arc $BM$}\) | \(=\) | \(\ds \text {arc $NC$}\) | Equal Arcs of Circles Subtended by Equal Straight Lines | |||||||||||
| \(\ds \text {arc $AM$}\) | \(=\) | \(\ds \text {arc $MC$}\) | by hypothesis | |||||||||||
| \(\ds \text {arc $AB$}\) | \(=\) | \(\ds \text {arc $MN$}\) | Common Notion $3$ | |||||||||||
| \(\ds AB\) | \(=\) | \(\ds MN\) | Equal Arcs of Circles Subtended by Equal Straight Lines | |||||||||||
| \(\ds AB\) | \(=\) | \(\ds DE\) | Common Notion $1$ | |||||||||||
| \(\ds AB + BC\) | \(=\) | \(\ds DE + EC\) | Common Notion $2$ | |||||||||||
| \(\ds AB + BC\) | \(=\) | \(\ds DE\) | addition |
The result follows.
$\blacksquare$