# Length of Angle Bisector

## 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:

$d^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$

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

## Proof 1

 $\displaystyle \frac {BD} {DC}$ $=$ $\displaystyle \frac c b$ Angle Bisector Theorem $\displaystyle \leadsto \ \$ $\displaystyle \frac {BD} {DC} + 1$ $=$ $\displaystyle \frac c b + 1$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {BD + DC} {DC}$ $=$ $\displaystyle \frac {b + c} b$ $\displaystyle \leadsto \ \$ $\displaystyle \frac a {DC}$ $=$ $\displaystyle \frac {b + c} b$ $\displaystyle \leadsto \ \$ $\displaystyle DC$ $=$ $\displaystyle \frac {a b} {b + c}$

Similarly, or by symmetry, we get:

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

From Stewart's Theorem, we have:

$b^2 \cdot BD + c^2 \cdot DC = d^2 \cdot a + BD \cdot DC \cdot a$

Substituting the above expressions for $BD$ and $DC$:

 $\displaystyle b^2 \dfrac {a c} {b + c} + c^2 \frac {a b} {b + c}$ $=$ $\displaystyle d^2 \cdot a + \dfrac {a c} {b + c} \cdot \frac {a b} {b + c} \cdot a$ $\displaystyle \leadsto \ \$ $\displaystyle a b c \frac {b + c} {b + c}$ $=$ $\displaystyle d^2 \cdot a + \frac{a^2 b c} {\paren {b + c}^2} \cdot a$ $\displaystyle \leadsto \ \$ $\displaystyle b c$ $=$ $\displaystyle d^2 + \frac {a^2 b c} {\paren {b + c}^2}$ $\displaystyle \leadsto \ \$ $\displaystyle d^2$ $=$ $\displaystyle b c \paren {1 - \frac {a^2} {\paren {b + c}^2} }$ $\displaystyle \leadsto \ \$ $\displaystyle d^2$ $=$ $\displaystyle \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$

$\blacksquare$

## Proof 2

From Length of Angle Bisector: Proof 1, we have:

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

Then we have:

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

Then from Triangles with Two Equal Angles are Similar we have:

$\triangle ABD \sim \triangle AFC$

So:

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

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

 $\displaystyle \frac c d$ $=$ $\displaystyle \frac {d + \frac {BD \cdot DC} d} b$ $\displaystyle \leadsto \ \$ $\displaystyle b c$ $=$ $\displaystyle d^2 + BD \cdot DC$ $\displaystyle \leadsto \ \$ $\displaystyle d^2$ $=$ $\displaystyle b c - BD \cdot DC$ $\displaystyle \leadsto \ \$ $\displaystyle d^2$ $=$ $\displaystyle b c - \frac {a c} {b + c} \cdot \frac {a b} {b + c}$ $\displaystyle \leadsto \ \$ $\displaystyle d^2$ $=$ $\displaystyle \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$

$\blacksquare$