Angles of Orthic Triangle of Acute Triangle/Proof
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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 angles of $\triangle DEF$ are $180 \degrees - 2 A$, $180 \degrees - 2 B$ and $180 \degrees - 2 C$.
Proof
Let $H$ be the orthocenter of $\triangle ABC$.
The quadrilateral $\Box FHDB$ is cyclic.
That is, $\Box FHDB$ can be circumscribed.
Hence:
\(\ds \angle HDF\) | \(=\) | \(\ds \angle HBF\) | Angles in Same Segment of Circle are Equal | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds 90 \degrees - A\) | as $\triangle AEB$ is a right triangle |
Similarly, $\Box DHEC$ is also a cyclic quadrilateral.
\(\ds \angle HDE\) | \(=\) | \(\ds \angle HCE\) | Angles in Same Segment of Circle are Equal | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds 90 \degrees - A\) | as $\triangle AFC$ is a right triangle |
Then:
\(\ds \angle FDE\) | \(=\) | \(\ds \angle HDE + \angle HDF\) | by construction | |||||||||||
\(\ds \) | \(=\) | \(\ds 180 \degrees - 2 A\) | adding $(1)$ and $(2)$ |
The same argument mutatis mutandis can be applied to $B$ and $C$.
Hence the result.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The pedal triangle