Euler Triangle Formula/Lemma 2
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 1
Without loss of generality, it will be demonstrated that $BP = IP$.
Let $CP$ be the bisector of $\angle ACB$.
- $\angle ACP$ and $\angle ICB$ both subtend $AP$.
Hence indirectly by the Inscribed Angle Theorem:
- $\angle ACP = \angle ICB$
From Angles on Equal Arcs are Equal:
- $\angle ACP = \angle ABP$
and so:
- $\angle ABP = \angle ICB$
By the construction of the incircle, $IB$ is the bisector of $B$.
Then:
- $\angle IBA = \angle IBC$
and so:
| \(\ds \angle IBP\) | \(=\) | \(\ds \angle IBA + \angle ABP\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \angle IBC + \angle ICB\) |
By Sum of Angles of Triangle equals Two Right Angles:
- $\angle CIB + \angle IBC + \angle ICB = \angle CIB + \angle PIB$
as $\angle CIB$ and $\angle PIB$ are supplementary.
Thus:
| \(\ds \angle PIB\) | \(=\) | \(\ds \angle IBC + \angle ICB\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \angle IBP\) |
and from Triangle with Two Equal Angles is Isosceles:
- $IP = BP$
The proof that $IP = AP$ follows the same lines.
$\blacksquare$
Proof 2
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
$\leadsto$:
- $AP = BP = IP$
$\blacksquare$