Distance from Vertex 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 $H$ be the orthocenter of $\triangle ABC$.

Then:

$AH = 2 R \cos A$

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:

\(\ds \angle ABH\) \(=\) \(\ds \angle ABE\)
\(\ds \) \(=\) \(\ds 180 \degrees - \angle AEB - \angle EAB\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds 180 \degrees - 90 \degrees - \angle EAB\) by construction
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds 90 \degrees - A\) simplifying


Consider $\Box EHCD$.

$\angle CDH$ 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:

\(\ds \angle AHB\) \(=\) \(\ds \angle EHD\) Two Straight Lines make Equal Opposite Angles
\(\ds \angle EHD + \angle DCE\) \(=\) \(\ds 180 \degrees\) Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \angle AHB\) \(=\) \(\ds 180 \degrees - C\)


Hence:

\(\ds \dfrac {AH} {\sin \angle ABH}\) \(=\) \(\ds \dfrac {AB} {\sin \angle AHB}\) Law of Sines applied to $\triangle ABH$
\(\ds \leadsto \ \ \) \(\ds \dfrac {AH} {\map \sin {90 \degrees - A} }\) \(=\) \(\ds \dfrac {AB} {\map \sin {180 \degrees - C} }\) from $(1)$ and $(2)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {AH} {\cos A}\) \(=\) \(\ds \dfrac {AB} {\sin C}\) Sine of Complement equals Cosine, Sine of Supplementary Angle
\(\ds \) \(=\) \(\ds 2 R\) Law of Sines applied to $\triangle ABC$

The result follows.

$\blacksquare$


Sources

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