Triangle is Orthic Triangle of Triangle formed from Excenters
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Theorem
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $I$ be the incenter of $\triangle ABC$.
Let $I_a$, $I_b$ and $I_c$ be the excenters of $\triangle ABC$ with respect to $a$, $b$ and $c$ respectively.
Let $\triangle I_a I_b I_c$ be the triangle whose vertices are those excenters of $I_a$, $I_b$ and $I_c$.
Then:
- $\triangle ABC$ is the orthic triangle of $\triangle I_a I_b I_c$
and:
- $I$ is the orthocenter of $\triangle I_a I_b I_c$.
Proof
From Construction of Excircle to Triangle, it is seen that:
- $A I_b$ is the angle bisector of $\angle PAC$
- $A I_c$ is the angle bisector of $\angle QAB$.
Hence $I_b A I_c$ is a straight line.
From the construction in Excenters and Incenter of Orthic Triangle, $A I I_a$ is a straight line which is perpendicular to $I_b A I_c$.
The same argument mutatis mutandis is used to show that:
- $B I I_b$ is a straight line which is perpendicular to the straight line $I_a B I_c$
and
- $C I I_c$ is a straight line which is perpendicular to the straight line $I_a C I_b$.
It follows by definition that :
- $\triangle ABC$ is the orthic triangle of $\triangle I_a I_b I_c$
and:
- $I$ is the orthocenter of $\triangle I_a I_b I_c$.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The ex-circles