Broken Chord Theorem/Proof 2
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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
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 |
$\blacksquare$