Angles of Orthic Triangle of Obtuse Triangle
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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 angles of $\triangle DEF$ are $2 A - 180 \degrees$, $2 B$ and $2 C$.
Proof
Let $H$ be the orthocenter of $\triangle ABC$.
The quadrilateral $\Box DBEA$ is cyclic.
That is, $\Box DBEA$ can be circumscribed.
Hence:
\(\ds \angle AED\) | \(=\) | \(\ds \angle ABD\) | Angles in Same Segment of Circle are Equal | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds B\) |
Similarly, $\Box BEFC$ is also a cyclic quadrilateral.
\(\ds \angle FEA\) | \(=\) | \(\ds \angle FBC\) | Angles in Same Segment of Circle are Equal | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds B\) |
Then:
\(\ds \angle DEF\) | \(=\) | \(\ds \angle AED + \angle FEA\) | by construction | |||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds 2 B\) | adding $(1)$ and $(2)$ |
The same argument mutatis mutandis can be applied to $C$, where we find:
- $(4): \quad \angle EFD = 2 C$
Hence:
\(\ds \angle EDF\) | \(=\) | \(\ds 180 \degrees - \angle EFD - \angle DEF\) | Sum of Angles of Triangle equals Two Right Angles | |||||||||||
\(\ds \) | \(=\) | \(\ds 180 \degrees - 2 \paren {B + C}\) | substituting from $(3)$ and $(4)$ and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 180 \degrees - 2 \paren {180 \degrees - A}\) | Sum of Angles of Triangle equals Two Right Angles | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 A - 180 \degrees\) | simplifying |
Hence the result.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(50)$