Tangent of Half Angle in Triangle/Proof

Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Let $s$ denote the semiperimeter of $\triangle ABC$:

$s = \dfrac {a + b + c} 2$

Then:

$\tan \dfrac C 2 = \sqrt {\dfrac {\paren {s - a} \paren {s - b} } {s \paren {s - c} } }$

$\ds \tan \dfrac C 2$

$=$

$\ds \dfrac {\sin \dfrac C 2} {\cos \dfrac C 2}$

$\ds $

$=$

$\ds \dfrac {\sqrt {\dfrac {\paren {s - a} \paren {s - b} } {a b} } } {\sqrt {\dfrac {s \paren {s - c} } {a b} } }$

$\ds $

$=$

$\ds \sqrt {\dfrac {\paren {s - a} \paren {s - b} } {s \paren {s - c} } }$

after simplification

$\blacksquare$