Length of Angle Bisector/Proof 1

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Theorem

Let $\triangle ABC$ be a triangle.

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

LengthOfAngleBisector.png

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

\(\ds \frac {BD} {DC}\) \(=\) \(\ds \frac c b\) Angle Bisector Theorem
\(\ds \leadsto \ \ \) \(\ds \frac {BD} {DC} + 1\) \(=\) \(\ds \frac c b + 1\)
\(\ds \leadsto \ \ \) \(\ds \frac {BD + DC} {DC}\) \(=\) \(\ds \frac {b + c} b\)
\(\ds \leadsto \ \ \) \(\ds \frac a {DC}\) \(=\) \(\ds \frac {b + c} b\)
\(\ds \leadsto \ \ \) \(\ds DC\) \(=\) \(\ds \frac {a b} {b + c}\)

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 $DC$:

\(\ds b^2 \dfrac {a c} {b + c} + c^2 \frac {a b} {b + c}\) \(=\) \(\ds d^2 \cdot a + \dfrac {a c} {b + c} \cdot \frac {a b} {b + c} \cdot a\)
\(\ds \leadsto \ \ \) \(\ds a b c \frac {b + c} {b + c}\) \(=\) \(\ds d^2 \cdot a + \frac{a^2 b c} {\paren {b + c}^2} \cdot a\)
\(\ds \leadsto \ \ \) \(\ds b c\) \(=\) \(\ds d^2 + \frac {a^2 b c} {\paren {b + c}^2}\)
\(\ds \leadsto \ \ \) \(\ds d^2\) \(=\) \(\ds b c \paren {1 - \frac {a^2} {\paren {b + c}^2} }\)
\(\ds \leadsto \ \ \) \(\ds d^2\) \(=\) \(\ds \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}\)

$\blacksquare$