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
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.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The orthocentre