Steiner-Lehmus Theorem/Proof 3
Theorem
Let $ABC$ be a triangle.
Denote the lengths of the angle bisectors through the vertices $A$ and $B$ by $\omega_\alpha$ and $\omega_\beta$.
Let $\omega_\alpha = \omega_\beta$.
Then $ABC$ is an isosceles triangle.
Proof
Draw $DF \parallel BE$ and $EF \parallel BD$.
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
- $\Box BEFD$ is a parallelogram.
- $\leadsto \angle DFE = \beta$
Draw $FA$.
Let $\angle EFA = \gamma$.
Let $\angle EAF = \delta$.
Aiming for a contradiction, suppose $\alpha > \beta$.
Compare $\triangle BAD$ with $\triangle ABE$.
The base is the same, $BA$.
| \(\ds BE\) | \(=\) | \(\ds AD\) | by hypothesis | |||||||||||
| \(\ds \alpha\) | \(>\) | \(\ds \beta\) | by hypothesis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds BD\) | \(>\) | \(\ds AE\) | Greater Angle of Triangle Subtended by Greater Side | ||||||||||
| \(\ds BD\) | \(=\) | \(\ds EF\) | Definition of Parallelogram | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds EF\) | \(>\) | \(\ds AE\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \delta\) | \(>\) | \(\ds \gamma\) | Greater Side of Triangle Subtends Greater Angle | ||||||||||
| \(\ds \alpha + \delta\) | \(>\) | \(\ds \beta + \gamma\) | Addition of Inequalities | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds AD\) | \(>\) | \(\ds DF\) | Greater Angle of Triangle Subtended by Greater Side | ||||||||||
| \(\ds DF\) | \(=\) | \(\ds BE\) | Definition of Parallelogram | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds AD\) | \(>\) | \(\ds BE\) |
But this is a contradiction.
Hence by Proof by Contradiction it cannot be the case that $\alpha > \beta$.
$\Box$
Aiming for a contradiction, suppose $\beta > \alpha$.
Again compare $\triangle BAD$ with $\triangle ABE$.
They have the same base, $BC$.
| \(\ds BE\) | \(=\) | \(\ds AD\) | by hypothesis | |||||||||||
| \(\ds \beta\) | \(>\) | \(\ds \alpha\) | by hypothesis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds AE\) | \(>\) | \(\ds BD\) | Greater Angle of Triangle Subtended by Greater Side | ||||||||||
| \(\ds BD\) | \(=\) | \(\ds EF\) | Definition of Parallelogram | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds AE\) | \(>\) | \(\ds EF\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \gamma\) | \(>\) | \(\ds \delta\) | Greater Side of Triangle Subtends Greater Angle | ||||||||||
| \(\ds \beta + \gamma\) | \(>\) | \(\ds \alpha + \delta\) | Addition of Inequalities | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds AD\) | \(>\) | \(\ds DF\) | Greater Angle of Triangle Subtended by Greater Side | ||||||||||
| \(\ds DF\) | \(=\) | \(\ds BE\) | Definition of Parallelogram | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds AD\) | \(>\) | \(\ds BE\) |
But this is a contradiction.
Hence by Proof by Contradiction it cannot be the case that $\alpha > \beta$.
$\Box$
Since neither $\alpha > \beta$ nor $\alpha < \beta$, it must be the case that
- $\alpha = \beta$
So by addition:
- $\angle ABC = \angle BAC$
and by Triangle with Two Equal Angles is Isosceles:
- $\triangle ABC$ is isosceles.
The result follows.
$\blacksquare$