Orthic Triangle of Obtuse Triangle

Theorem
Let $\triangle ABC$ be an obtuse triangle such that $A$ is the obtuse angle.

Let $H$ be the orthocenter of $\triangle ABC$.

Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.

Then $\triangle DEF$ is also the orthic triangle of $\triangle HBC$, which is an acute triangle.

Proof

 * Orthic-Triangle-Obtuse.png

By construction:
 * $CE \perp BH$
 * $BF \perp CH$
 * $HD \perp BC$

Thus by definition $CE$, $BF$ and $HD$ are the altitudes of $\triangle HBC$.

Also by construction, $A$ lies on $CE$, $BF$ and $HD$.

Hence $\triangle DEF$ is the orthic triangle of $\triangle HBC$.