Angles of Orthic Triangle of Acute Triangle

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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

Orthic-Triangle.png

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