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

Orthic-Triangle-Obtuse.png

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