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Angle Bisector Theorem

Angle Bisector Theorem

Lesson · Beginner

Geometry: Plane Geometry

Angle Bisector Theorem

\frac{BD}{DC}=\frac{c}{b}

Triangle Proportionality Theorem

Angle Bisector Theorem proof

Draw a line through $B$ parallel to $AD$ , intersecting the extension of $AC$ at $E$ , as shown.

Angle Bisector Theorem proof

Since $AD \parallel BE$

\angle AEB=\angle CAD=\frac{A}{2}

\angle ABE=\angle BAD=\frac{A}{2}

Therefore

AE=AB=c

By the Triangle Proportionality Theorem (Thales' Theorem)

\frac{CD}{DB}=\frac{CA}{AE}=\frac{b}{c}

Therefore

\frac{BD}{DC}=\frac{c}{b}

Angle Bisector Theorem

\frac{BD}{CD}=\frac{c}{b}\Rightarrow\frac{BD}{BD+CD}=\frac{c}{b+c}

\frac{BD}{a}=\frac{c}{b+c}

BD=\frac{ac}{b+c}

Similarly

CD=\frac{ab}{b+c}

∠A = 60°, AD is the angle bisector of A. AB = 3, AC = 4. Find AD.

Angle Bisector Theorem Example 1

Angle Bisector Theorem Example 1 solution

BD=\frac{ac}{b+c}=\frac{3a}{7}

By law of sines in triangle ABD

\frac{AD}{\sin B}=\frac{BD}{\sin30^\circ}=\frac{\frac{3a}{7}}{\frac12}=\frac{6a}{7}

AD=\frac{6a\sin B}{7}\qquad (1)

By law of sines in triangle ABC

\frac{a}{\sin60^\circ}=\frac{b}{\sin B}

a\sin B=b\sin60^\circ=\frac{4\sqrt3}{2}=2\sqrt3\qquad (2)

By (1), (2)

AD=\frac67\cdot2\sqrt3=\frac{12\sqrt3}{7}

AMC 10A/12A 2025 (Problem 23/16)
AMC 12A 2025 (Problem 8)

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