Euler Triangle Formula/Lemma 2/Proof 1

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

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.

$\blacksquare$