Distance from Foot of Altitude of Triangle to Orthocenter/Proof

Proof

 * Vertex-to-Orthocenter.png

We construct the circumcircle of $\triangle ABC$, whose circumcenter is $K$ and whose circumradius is $R$.

We construct the orthocenter $H$ of $\triangle ABC$ as the intersection of the altitudes $AD$ and $BE$.

First we note the following:

$\angle CDH$ and $\angle CEH$ are both right angles.

Hence from Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\Box EHCD$ is a cyclic quadrilateral.

Thus:

Hence:

Then we have:

The result follows.