Altitude of a Triangle

The distance between a vertex of a triangle and the opposite side is an altitude. Formally, the shortest line segment between a vertex of a triangle and the (possibly extended) opposite side.

Note: Every triangle have 3 altitudes which intersect at one point called the orthocenter.

Altitudes of an acute triangle

The orthocenter is an interior point for the triangle.

∠ AHB = 180 - γ = α + β
∠ BHC = 180 - α = β + γ
∠ AHC = 180 - β = α + γ
∠ AHHc = β, ∠ BHHc = α, ∠ BHHa = γ

Altitudes of an obtuse triangle

The orthocenter lies outside the triangle.
Also two of the altitudes always lies outside the triangle.
∠ AHHc = ∠ CBA = β
∠ HcHB = ∠ CAB = α

Right triangle

The altitude AHa coincide AC.
The altitude BHb coincide BC.
The orthocenter H coincide C.
∠ ACHc = β, ∠ BCHc

Formulas

$AH_a:BH_b:CH_c=\frac{1}{a}:\frac{1}{b}:\frac{1}{c}$

$\frac{a}{AH_a}=\frac{b}{BH_b}=\frac{c}{\frac{AH_aBH_b}{CH_c}}$

R - the radius of circumscribed circle
r - the radius of inscribed circle
p - is half perimeter: (a + b + c)/2

$AH_a=b \sin\gamma=c \sin\beta=\frac{a \sin\beta \sin\gamma}{\sin\alpha}=$

$=2R \sin\beta\ \sin\gamma=\frac{bc}{2R}=\frac{2\sqrt{p(p-a)(p-b)(p-c)}}{a}$

$BH_b=a\ \sin\gamma=c\ \sin\alpha=\frac{b\ \sin\alpha\ \sin\gamma}{\sin\beta}=$
$=2R\ \sin\alpha \sin \gamma=\frac{ac}{2R}=\frac{2\sqrt{p(p-a)(p-b)(p-c)}}{b}$

$CH_c=a\ \sin\beta=b\ \sin\alpha=\frac{c\ \sin\alpha\ \sin\beta}{\sin\gamma}=$
$=2R\ \sin\alpha \sin \beta=\frac{ab}{2R}=\frac{2\sqrt{p(p-a)(p-b)(p-c)}}{c}$

$\frac{1}{AH_a}+\frac{1}{BH_b}+\frac{1}{CH_c}=\frac{1}{r}$

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