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

Na geometria, 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.

Conteúdo 1 Teorema 2 Provas 2.1 Prova 1 2.2 Prova 2 2.3 Prova 3 3 Exterior angle bisectors 4 História 5 Formulários 6 Referências 7 Leitura adicional 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: {estilo de exibição {fratura {|BD|}{|CD|}}={fratura {|AB|}{|CA|}},} e inversamente, 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, então {estilo de exibição {fratura {|BD|}{|CD|}}={fratura {|AB|sin angle DAB}{|CA|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: {estilo de exibição {fratura {|AB|}{|BD|}}={fratura {sin angle ADB}{sin angle DAB}}} (1) {estilo de exibição {fratura {|CA|}{|CD|}}={fratura {sin angle ADC}{sin angle DAC}}} (2) Angles ∠ ADB and ∠ ADC form a linear pair, isso é, they are adjacent supplementary angles. Since supplementary angles have equal sines, {estilo de exibição {sin angle ADB}={sin angle ADC}.} Angles ∠ DAB and ∠ DAC are equal. Portanto, the right hand sides of equations (1) e (2) are equal, so their left hand sides must also be equal.

{estilo de exibição {fratura {|BD|}{|CD|}}={fratura {|AB|}{|CA|}},} which is the angle bisector theorem.

If angles ∠ DAB and ∠ DAC are unequal, equations (1) e (2) can be re-written as: {estilo de exibição {{fratura {|AB|}{|BD|}}sin angle DAB=sin angle ADB},} {estilo de exibição {{fratura {|CA|}{|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: {estilo de exibição {{fratura {|AB|}{|BD|}}sin angle DAB={fratura {|CA|}{|CD|}}sin angle DAC},} which rearranges to the "generalizado" version of the theorem.

Prova 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. Então, 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 (isso é, between B and C) and they are identical in the other cases being considered, so the triangles DB1B and DC1C are similar (AAA), o que implica que: {estilo de exibição {fratura {|BD|}{|CD|}}={fratura {|BB_{1}|}{|CC_{1}|}}={fratura {|AB|sin angle BAD}{|CA|sin angle CAD}}.} If D is the foot of an altitude, então, {estilo de exibição {fratura {|BD|}{|AB|}}=sin angle BAD{texto{ e }}{fratura {|CD|}{|CA|}}=sin angle DAC,} and the generalized form follows.

Prova 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} e {displaystyle triangle CAD} , which are created by the angle bisector in {estilo de exibição A} . Computing those areas twice using different formulas, isso é {estilo de exibição {tfrac {1}{2}}gh} with base {estilo de exibição g} and altitude {estilo de exibição h} e {estilo de exibição {tfrac {1}{2}}absin(gama )} with sides {estilo de exibição a} , {estilo de exibição b} and their enclosed angle {gama de estilo de exibição } , will yield the desired result.