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

Orthic-Triangle-Obtuse.png

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