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In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC:

$${\frac {|BD|} {|DC|}}={\frac {|AB|}{|AC|}}.$$

The generalized angle bisector theorem states that if D lies on BC, then

$${\frac {|BD|} {|DC|}}={\frac {|AB| \sin \angle DAB}{|AC| \sin \angle DAC}}.$$

This reduces to the previous version if AD is the bisector of BAC.

The angle bisector theorem is commonly used when the angle bisectors and side lengths are known. It can be used in a calculation or in a proof.

An angle bisector of an isosceles triangle will also bisect the opposite side, when the angle bisect bisects the vertex angle of the triangle
Proof

In the above diagram, use the Law of Sines on triangles ABD and ACD:

$${\frac {|AB|} {|BD|}} = {\frac {\sin \angle BDA} {\sin \angle BAD}} .....$$ (Equation 1)

$${\frac {|AC|} {|DC|}} = {\frac {\sin \angle ADC} {\sin \angle DAC}} .....$$ (Equation 2)

Angles BDA and ADC are supplementary, and therefore have equal sines. Angles BAD and DAC are equal. Therefore the Right Hand Sides of Equations 1 and 2 are equal, so their Left Hand Sides must also be equal:

$${\frac {|AB|} {|BD|}}={\frac {|AC|}{|DC|}}$$

which is the Angle Bisector Theorem.

If angles BAD and DAC are unequal, Equations 1 and 2 can be re-written as:

$${\frac {|AB|} {|BD|} \sin \angle\ BAD = \sin \angle BDA}$$

$${\frac {|AC|} {|DC|} \sin \angle\ DAC = \sin \angle ADC} Angles BDA and ADC are still supplementary, so the Right Hand Sides of these equations are equal, so the Left Hand Sides are equal to: \( {\frac {|AB|} {|BD|} \sin \angle\ BAD = \frac {|AC|} {|DC|} \sin \angle\ DAC}$$

which rearranges to the "generalized" version of the theorem.

An alternative proof goes as follows, using its own diagram:

Let B1 be the base of altitude in the triangle ABD through B and let C1 be the base of altitude in the triangle ACD through C. Then,

DB1B and DC1C are right, while the angles B1DB and C1DC are congruent if D lies on the segment BC and they are identical otherwise, so the triangles DB1B and DC1C are similar (AAA), which implies that

$${\frac {|BD|} {|CD|}}= {\frac {|BB_1|}{|CC_1|}}=\frac {|AB|\sin \angle BAD}{|AC|\sin \angle CAD}.$$