Length of Angle Bisector/Proof 2

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Theorem

Let $\triangle ABC$ be a triangle.

Let $AD$ be the angle bisector of $\angle BAC$ in $\triangle ABC$.

Let $d$ be the length of $AD$.

Then $d$ is given by:

$d^2 = \dfrac {b c} {\left({b + c}\right)^2} \left({\left({b + c}\right)^2 - a^2}\right)$

where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.

$BD = \dfrac {a c} {b + c}$

$DC = \dfrac {a b} {b + c}$

Then we have:

	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \angle BAD$	$\cong$	$\displaystyle \angle FAC$	$\displaystyle $	$\displaystyle $	$\displaystyle $	angle bisector
	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \angle ABD$	$\cong$	$\displaystyle \angle AFC$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Angles in Same Segment of Circle are Equal

$\triangle ABD \sim \triangle AFC$

So:

	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \frac c d$	$=$	$\displaystyle \frac {AF} b$	$\displaystyle $	$\displaystyle $	$\displaystyle $	as $\triangle ABD$ and $\triangle AFC$ are similar
	$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \frac c d$	$=$	$\displaystyle \frac {d + DF} b$	$\displaystyle $	$\displaystyle $	$\displaystyle $

Now we use the Intersecting Chord Theorem, which gives us $BD \cdot DC = d \cdot DF$.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \frac c d$	$=$	$\displaystyle \frac {d + \frac {BD \cdot DC} d} b$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle b c$	$=$	$\displaystyle d^2 + BD \cdot DC$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle d^2$	$=$	$\displaystyle b c - BD \cdot DC$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle d^2$	$=$	$\displaystyle b c - \frac {a c} {b + c} \cdot \frac {a b} {b + c}$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle d^2$	$=$	$\displaystyle \frac {b c}{\left({b + c}\right)^2}\left({\left({b + c}\right)^2 - a^2}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$\blacksquare$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle BAD\)	\(\cong\)	\(\displaystyle \angle FAC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	angle bisector
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle ABD\)	\(\cong\)	\(\displaystyle \angle AFC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Angles in Same Segment of Circle are Equal

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac c d\)	\(=\)	\(\displaystyle \frac {AF} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $\triangle ABD$ and $\triangle AFC$ are similar
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac c d\)	\(=\)	\(\displaystyle \frac {d + DF} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac c d\)	\(=\)	\(\displaystyle \frac {d + \frac {BD \cdot DC} d} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle b c\)	\(=\)	\(\displaystyle d^2 + BD \cdot DC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle d^2\)	\(=\)	\(\displaystyle b c - BD \cdot DC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle d^2\)	\(=\)	\(\displaystyle b c - \frac {a c} {b + c} \cdot \frac {a b} {b + c}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle d^2\)	\(=\)	\(\displaystyle \frac {b c}{\left({b + c}\right)^2}\left({\left({b + c}\right)^2 - a^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)