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 $\triangle DEF$ $EF$ is $A$
 * the excenter of $\triangle DEF$ $DF$ is $B$
 * the excenter of $\triangle DEF$ $DE$ is $C$

and:
 * the incenter of $\triangle DEF$ is the orthocenter of $\triangle ABC$.

Proof

 * Excircle-of-Orthic-Triangle.png

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 $\triangle DEF$ $DE$.

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


 * Orthic-Triangle.png

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 $\triangle DEF$.