Distance from Foot of Altitude of Triangle to Orthocenter/Proof

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Theorem

Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Let $D$ be the foot of the altitude of $\angle A$

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

Then:

$HD = 2 R \cos B \cos C$

where $R$ denotes the circumradius of $\triangle ABC$.


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:

\(\ds \angle EHD + \angle DCE\) \(=\) \(\ds 180 \degrees\) Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles
\(\ds \angle BHD + \angle EHD\) \(=\) \(\ds 180 \degrees\)
\(\ds \leadsto \ \ \) \(\ds \angle BHD\) \(=\) \(\ds C\)
\(\ds \leadsto \ \ \) \(\ds \angle HBD\) \(=\) \(\ds 90 \degrees - C\)


Hence:

\(\ds \dfrac {HD} {HB}\) \(=\) \(\ds \sin \angle HBD\) Definition of Sine of Angle
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds HD\) \(=\) \(\ds HB \cos C\) Sine of Complement equals Cosine


Then we have:

\(\ds HB\) \(=\) \(\ds 2 R \cos B\) Distance from Vertex of Triangle to Orthocenter
\(\ds \leadsto \ \ \) \(\ds HD\) \(=\) \(\ds 2 R \cos B \cos C\) from $(1)$

The result follows.

$\blacksquare$


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