Euler Triangle Formula/Lemma 2/Proof 2
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Lemma to Euler Triangle Formula
Let the bisector of angle $C$ of triangle $\triangle ABC$ be produced to the circumcircle at $P$.
Let $I$ be the incenter of $\triangle ABC$.
Then:
- $AP = BP = IP$
Proof
We have by hypothesis:
Let the half-angles be:
- $\alpha = \dfrac 1 2 \angle CAB$
- $\beta = \dfrac 1 2 \angle ABC$
- $\gamma = \dfrac 1 2 \angle BCA$
| \(\ds AP\) | \(=\) | \(\ds BP\) | Equal Angles in Equal Circles | |||||||||||
| \(\ds \angle BAP\) | \(=\) | \(\ds \gamma\) | Angles on Equal Arcs are Equal | |||||||||||
| \(\ds \angle API\) | \(=\) | \(\ds \angle ABC = 2 \beta\) | Angles on Equal Arcs are Equal | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \angle IAP\) | \(=\) | \(\ds \alpha + \gamma\) | |||||||||||
| \(\ds \angle AIP\) | \(=\) | \(\ds \alpha + \gamma\) | Sum of Angles of Triangle equals Two Right Angles |
By Triangle with Two Equal Angles is Isosceles
- $\triangle AIP$ is isosceles
$\leadsto$:
- $AP = BP = IP$
$\blacksquare$