Sides of Orthic Triangle of Obtuse Triangle/Proof
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Theorem
Let $\triangle ABC$ be an obtuse triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $\angle A$ be the obtuse angle of $\triangle ABC$.
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 $\triangle ABC$.
\(\ds \frac {EF} {\sin \angle EAF}\) | \(=\) | \(\ds \frac {AF} {\sin \angle AEF}\) | Law of Sines for $\triangle AFE$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {EF} {\sin A}\) | \(=\) | \(\ds \frac {b \map \cos {180 \degrees - A} } {\sin B}\) | Angles in Same Segment of Circle are Equal: $BEFC$ is cyclic | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a \map \cos {180 \degrees - A} } {\sin A}\) | Law of Sines for $\triangle ABC$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds EF\) | \(=\) | \(\ds a \map \cos {180 \degrees - A}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -a \cos A\) | Cosine of Supplementary Angle |
On the other hand,
\(\ds \frac {DE} {\sin \angle DAE}\) | \(=\) | \(\ds \frac {AD} {\sin \angle AED}\) | Law of Sines for $\triangle ADE$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {DE} {\map \sin {90 \degrees + C} }\) | \(=\) | \(\ds \frac {b \sin C} {\sin B}\) | Angles in Same Segment of Circle are Equal: $BEAD$ is cyclic | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {c \sin C} {\sin C}\) | Law of Sines for $\triangle ABC$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds DE\) | \(=\) | \(\ds c \map \sin {90 \degrees + C}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds c \cos C\) | Sine of Angle plus Right Angle |
Similar arguments mutatis mutandis can be applied to $DF$ by swapping $E$ with $F$ and $B$ with $C$.
Hence the result.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The pedal triangle