Euler Triangle Formula/Lemma

Lemma to Euler Triangle Formula

 * EulerTriangleLemma.png

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
, it will be demonstrated that $BP = IP$.

Let $CP$ be the bisector of $\angle ACB$.

We have therefore that:
 * $\angle ACP = \angle ICB$

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:

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:

and from Triangle with Two Equal Angles is Isosceles:
 * $IP = BP$

The proof that $IP = AP$ follows the same lines.