Broken Chord Theorem/Proof 5
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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
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.
$\blacksquare$