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

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 $\triangle HBC$.

This needs considerable tedious hard slog to complete it. In particular: Demonstrate that $\triangle HBC$ is acute To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.