Length of Angle Bisector in terms of Angle

Theorem

Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Let $AD$ be the angle bisector of $\angle BAC$ that intersects $a$ at $D$.

Then:

$AD = \dfrac {2 c b \cos \frac A 2} {b + c}$

Proof

\(\ds \frac {BD} {DC}\)

\(=\)

\(\ds \frac c b\)

Angle Bisector Theorem

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds BD\)

\(=\)

\(\ds \frac {a c} {b + c}\)

as $a = BD + DC$

Then:

\(\ds \frac {AD} {\sin B}\)

\(=\)

\(\ds \frac {BD} {\sin \frac A 2}\)

applying Law of Sines to $\triangle ADB$

\(\ds \)

\(=\)

\(\ds \frac {a c} {\paren {b + c} \sin \frac A 2}\)

substituting for $BD$ from $(1)$

\(\ds \leadsto \ \ \)

\(\ds AD\)

\(=\)

\(\ds \frac {a c \sin B} {\paren {b + c} \sin \frac A 2}\)

rearranging

\(\ds \)

\(=\)

\(\ds \frac {c b \sin A} {\paren {b + c} \sin \frac A 2}\)

Law of Sines

\(\ds \)

\(=\)

\(\ds \frac {2 c b \sin \frac A 2 \cos \frac A 2} {\paren {b + c} \sin \frac A 2}\)

Double Angle Formula for Sine

\(\ds \)

\(=\)

\(\ds \frac {2 c b \cos \frac A 2} {b + c}\)

simplifying

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The angle bisector