Sides of Orthic Triangle of Acute Triangle/Proof

Theorem

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

Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.

Then the sides of $\triangle DEF$ are $a \cos A$, $b \cos B$ and $c \cos C$.

Proof

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

Let $R$ be the circumradius of $\triangle ABC$.

\(\ds \dfrac {EF} {\sin A}\)

\(=\)

\(\ds \dfrac {AE} {\sin \angle AFE}\)

Law of Sines for $\triangle AFE$

\(\ds \)

\(=\)

\(\ds \dfrac {c \cos A} {\sin C}\)

\(\ds \)

\(=\)

\(\ds 2 R \cos A\)

\(\ds \leadsto \ \ \)

\(\ds EF\)

\(=\)

\(\ds 2 R \cos A \sin A\)

\(\ds \)

\(=\)

\(\ds a \cos A\)

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The same argument mutatis mutandis can be applied to $DF$ and $DE$.

Hence the result.

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The pedal triangle