Angle bisector theorem

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.

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'':
:$=,$

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

:$=.$

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'':

Angles ''∠ ADB'' and ''∠ ADC'' form a linear pair, that is, they are adjacent supplementary angles. Since supplementary angles have equal sines,

:$= .$

Angles ''∠ DAB'' and ''∠ DAC'' are equal. Therefore, the right hand sides of equations () and () are equal, so their left hand sides must also be equal.

:$=,$

which is the angle bisector theorem.

If angles ''∠ DAB'' and ''∠ DAC'' are unequal, equations () and () can be re-written as:

:$,$
:$.$

Angles ''∠ ADB'' and ''∠ ADC'' are still supplementary, so the right hand sides of these equations are still equal, so we obtain:

:$,$

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 ''B''₁ be the base (foot) of the altitude in the triangle ''ABD'' through ''B'' and let ''C''₁ 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 ''B''₁ or ''C''₁ lies inside triangle ''ABC'' and it can be assumed without loss of generality that ''B''₁ does. This case is depicted in the adjacent diagram. If ''D'' lies outside of segment ''BC'', then neither ''B''₁ nor ''C''₁ lies inside the triangle.

''∠ DB''₁''B'' and ''∠ DC''₁''C'' are right angles, while the angles ''∠ B''₁''DB'' and ''∠ C''₁''DC'' 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 ''DB''₁''B'' and ''DC''₁''C'' are similar (AAA), which implies that:

:$= = \frac .$

If ''D'' is the foot of an altitude, then,
:$\frac = \sin \angle \ BAD \text \frac = \sin \angle \ DAC,$
and the generalized form follows.

Proof 3

A quick proof can be obtained by looking at the ratio of the areas of the two triangles $\triangle BAD$ and $\triangle CAD$, which are created by the angle bisector in $A$. Computing those areas twice using different formulas, that is $\fracgh$ with base $g$ and altitude $h$ and $\fracab\sin(\gamma)$ with sides $a$, $b$ and their enclosed angle $\gamma$, will yield the desired result.

Let $h$ denote the height of the triangles on base $BC$ and $\alpha$ be half of the angle in $A$. Then
:$\frac = \frac = \frac$

and
:$\frac = \frac = \frac$

yields
:$\frac = \frac.$

Exterior angle bisectors

For the exterior angle bisectors in a non-equilateral triangle there exist similar equations for the ratios of the lengths of triangle sides. More precisely if the exterior angle bisector in $A$ intersects the extended side $BC$ in $E$, the exterior angle bisector in $B$ intersects the extended side $AC$ in $D$ and the exterior angle bisector in $C$ intersects the extended side $AB$ in $F$, then the following equations hold:

:$\frac = \frac$, $\frac = \frac$, $\frac = \frac$

The three points of intersection between the exterior angle bisectors and the extended triangle sides $D$, $E$ and $F$ are collinear, that is they lie on a common line.

History

The angle bisector theorem appears as Proposition 3 of Book VI in Euclid's Elements. According to , the corresponding statement for an external angle bisector was given by Robert Simson

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