Excenters and Incenter of Orthic Triangle/Acute Triangle

Theorem

Let $\triangle ABC$ be an acute triangle.

Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ such that:

$D$ is on $BC$

$E$ is on $AC$

$F$ is on $AB$

Then:

the excenter of $\triangle DEF$ with respect to $EF$ is $A$

the excenter of $\triangle DEF$ with respect to $DF$ is $B$

the excenter of $\triangle DEF$ with respect to $DE$ is $C$

and:

the incenter of $\triangle DEF$ is the orthocenter of $\triangle ABC$.

Proof

From Altitudes of Triangle Bisect Angles of Orthic Triangle, $AD$ is the angle bisector of $\angle FDE$.

From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, the angle bisector of $\angle PDE$ is perpendicular to $AD$.

The line perpendicular to $AD$ is $BC$.

Similarly, from Altitudes of Triangle Bisect Angles of Orthic Triangle, $BE$ is the angle bisector of $\angle FED$.

From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, the angle bisector of $\angle DEQ$ is perpendicular to $BE$.

The line perpendicular to $BE$ is $AC$.

From Construction of Excircle to Triangle, the intersection of $AC$ and $BC$ is the excenter of $\triangle DEF$ with respect to $DE$.

The same argument can be used mutatis mutandis to demonstrate the locations of the excenter of $\triangle DEF$ with respect to $DF$ and $EF$.

$\Box$

From Altitudes of Triangle Bisect Angles of Orthic Triangle:

$AD$ is the angle bisector of $\angle FDE$

$BE$ is the angle bisector of $\angle DEF$

$FC$ is the angle bisector of $\angle EFD$

From Line from Vertex of Triangle to Incenter is Angle Bisector it follows that $H$ is the incenter of $\triangle DEF$.

$\blacksquare$