# Angle bisector theorem

Angle bisector theorem In this diagram, BD:DC = AB:AC.

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.

Contents 1 Theorem 2 Proofs 2.1 Proof 1 2.2 Proof 2 2.3 Proof 3 3 Exterior angle bisectors 4 History 5 Applications 6 References 7 Further reading 8 External links Theorem Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D between B and C. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment CD is equal to the ratio of the length of side AB to the length of side AC: {displaystyle {frac {|BD|}{|CD|}}={frac {|AB|}{|AC|}},} and conversely, if a point D on the side BC of triangle ABC divides BC in the same ratio as the sides AB and AC, then AD is the angle bisector of angle ∠ A.

The generalized angle bisector theorem states that if D lies on the line BC, then {displaystyle {frac {|BD|}{|CD|}}={frac {|AB|sin angle DAB}{|AC|sin angle DAC}}.} This reduces to the previous version if AD is the bisector of ∠ BAC. When D is external to the segment BC, directed line segments and directed angles must be used in the calculation.

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 immediate consequence of the theorem is that the angle bisector of the vertex angle of an isosceles triangle will also bisect the opposite side.

Proofs Proof 1 In the above diagram, use the law of sines on triangles ABD and ACD: {displaystyle {frac {|AB|}{|BD|}}={frac {sin angle ADB}{sin angle DAB}}}         (1) {displaystyle {frac {|AC|}{|CD|}}={frac {sin angle ADC}{sin angle DAC}}}         (2) Angles ∠ ADB and ∠ ADC form a linear pair, that is, they are adjacent supplementary angles. Since supplementary angles have equal sines, {displaystyle {sin angle ADB}={sin angle ADC}.} Angles ∠ DAB 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.

{displaystyle {frac {|BD|}{|CD|}}={frac {|AB|}{|AC|}},} which is the angle bisector theorem.

If angles ∠ DAB and ∠ DAC are unequal, equations (1) and (2) can be re-written as: {displaystyle {{frac {|AB|}{|BD|}}sin angle DAB=sin angle ADB},} {displaystyle {{frac {|AC|}{|CD|}}sin angle DAC=sin angle ADC}.} Angles ∠ ADB and ∠ ADC are still supplementary, so the right hand sides of these equations are still equal, so we obtain: {displaystyle {{frac {|AB|}{|BD|}}sin angle DAB={frac {|AC|}{|CD|}}sin angle DAC},} which rearranges to the "generalized" version of the theorem.

Proof 2 Let D be a point on the line BC, not equal to B or C and such that AD is not an altitude of triangle ABC.

Let B1 be the base (foot) of the altitude in the triangle ABD through B and let C1 be the base of the altitude in the triangle ACD through C. Then, if D is strictly between B and C, one and only one of B1 or C1 lies inside triangle ABC and it can be assumed without loss of generality that B1 does. This case is depicted in the adjacent diagram. If D lies outside of segment BC, then neither B1 nor C1 lies inside the triangle.

∠ DB1B and ∠ DC1C are right angles, while the angles ∠ B1DB and ∠ C1DC are congruent if D lies on the segment BC (that is, between B and C) and they are identical in the other cases being considered, so the triangles DB1B and DC1C are similar (AAA), which implies that: {displaystyle {frac {|BD|}{|CD|}}={frac {|BB_{1}|}{|CC_{1}|}}={frac {|AB|sin angle BAD}{|AC|sin angle CAD}}.} If D is the foot of an altitude, then, {displaystyle {frac {|BD|}{|AB|}}=sin angle BAD{text{ and }}{frac {|CD|}{|AC|}}=sin angle DAC,} and the generalized form follows.

Proof 3 {displaystyle alpha ={tfrac {angle BAC}{2}}=angle BAD=angle CAD} A quick proof can be obtained by looking at the ratio of the areas of the two triangles {displaystyle triangle BAD} and {displaystyle triangle CAD} , which are created by the angle bisector in {displaystyle A} . Computing those areas twice using different formulas, that is {displaystyle {tfrac {1}{2}}gh} with base {displaystyle g} and altitude {displaystyle h} and {displaystyle {tfrac {1}{2}}absin(gamma )} with sides {displaystyle a} , {displaystyle b} and their enclosed angle {displaystyle gamma } , will yield the desired result.