Altitudes of Triangle Bisect Angles of Orthic Triangle

Theorem

Let $\triangle ABC$ be a triangle.

Let $\triangle DEF$ be its orthic triangle.

The altitudes of $\triangle ABC$ are the angle bisectors of $\triangle DEF$.

Consider the triangles $\triangle ABE$ and $\triangle ACF$.

We have that:

$\angle FAC$ and $\angle BAE$ are common

$\angle AFC$ and $\angle AEB$ are both right angles

and it follows from Triangles with Two Equal Angles are Similar that $\triangle ABE$ and $\triangle ACF$ are similar.

Thus:

$\angle ABE = \angle ACF$

Consider the quadrilateral $\Box BFHD$.

We have that $\angle BFH$ and $\angle BDH$ are both right angles.

Thus two opposite angles of $\Box BFHD$ sum to two right angles

$\angle FBH = \angle FDH$.

By similar analysis of quadrilateral $\Box DHEC$, we note that:

$\angle HDE = \angle HCE$

But then we have:

$\angle FBH = \angle ABE$

and:

$\angle HCE = \angle ACF$

Hence it follows that:

$\angle FDH = \angle HDE$

demonstrating that $AD$ is the angle bisector of $\angle FDE$.

The same argument applies mutatis mutandis to $\angle FDE$ and $\angle FED$.

Hence the result.

$\blacksquare$