Length of Angle Bisector
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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:
- $d^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.
Proof 1
\(\ds \frac {BD} {DC}\) | \(=\) | \(\ds \frac c b\) | 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 $DC$:
\(\ds b^2 \dfrac {a c} {b + c} + c^2 \frac {a b} {b + c}\) | \(=\) | \(\ds d^2 \cdot a + \dfrac {a c} {b + c} \cdot \frac {a b} {b + c} \cdot a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a b c \frac {b + c} {b + c}\) | \(=\) | \(\ds d^2 \cdot a + \frac{a^2 b c} {\paren {b + c}^2} \cdot a\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b c\) | \(=\) | \(\ds d^2 + \frac {a^2 b c} {\paren {b + c}^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds d^2\) | \(=\) | \(\ds b c \paren {1 - \frac {a^2} {\paren {b + c}^2} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds d^2\) | \(=\) | \(\ds \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}\) |
$\blacksquare$
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:
\(\ds \angle BAD\) | \(\cong\) | \(\ds \angle FAC\) | Definition of Angle Bisector | |||||||||||
\(\ds \angle ABD\) | \(\cong\) | \(\ds \angle AFC\) | 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:
\(\ds \frac c d\) | \(=\) | \(\ds \frac {AF} b\) | as $\triangle ABD$ and $\triangle AFC$ are similar | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac c d\) | \(=\) | \(\ds \frac {d + DF} b\) |
Now we use the Intersecting Chords Theorem, which gives us $BD \cdot DC = d \cdot DF$.
\(\ds \frac c d\) | \(=\) | \(\ds \frac {d + \frac {BD \cdot DC} d} b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds b c\) | \(=\) | \(\ds d^2 + BD \cdot DC\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds d^2\) | \(=\) | \(\ds b c - BD \cdot DC\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds d^2\) | \(=\) | \(\ds b c - \frac {a c} {b + c} \cdot \frac {a b} {b + c}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds d^2\) | \(=\) | \(\ds \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}\) |
$\blacksquare$