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 . Let the angle bisector of angle intersect side at a point between and . The angle bisector theorem states that the ratio of the length of the line segment to the length of segment is equal to the ratio of the length of side to the length of side :
:$=,$

and conversely, if a point on the side of divides in the same ratio as the sides and , then is the angle bisector of angle .

The generalized angle bisector theorem (which is not necessarily an angle bisector theorem, since the angle is not necessarily bisected into equal parts) states that if lies on the line , then

:$=.$

This reduces to the previous version if is the bisector of . When is external to the segment , 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

There exist many different ways of proving the angle bisector theorem. A few of them are shown below.

Proof using similar triangles

As shown in the accompanying animation, the theorem can be proved using similar triangles. In the version illustrated here, the triangle $\triangle ABC$ gets reflected across a line that is perpendicular to the angle bisector $AD$, resulting in the triangle $\triangle A B_2 C_2$ with bisector $AD_2$. The fact that the bisection-produced angles $\angle BAD$ and $\angle CAD$ are equal means that $BA C_2$ and $CA B_2$ are straight lines. This allows the construction of triangle $\triangle C_2BC$ that is similar to $\triangle ABD$. Because the ratios between corresponding sides of similar triangles are all equal, it follows that $, AB, /, AC_2, = , BD, /, CD,$. However, $AC_2$ was constructed as a reflection of the line $AC$, and so those two lines are of equal length. Therefore, $, AB, /, AC, = , BD, /, CD,$, yielding the result stated by the theorem.

Proof using law of sines

In the above diagram, use the law of sines on triangles and :

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

:$= .$

Angles and 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 are unequal, equations () and () can be re-written as:

:$,$
:$.$

Angles 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 using triangle altitudes

Let be a point on the line , not equal to or and such that is not an altitude of triangle .

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

are right angles, while the angles are congruent if lies on the segment (that is, between and ) and they are identical in the other cases being considered, so the triangles are similar (AAA), which implies that:

:$= = \frac .$

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

Proof using triangle areas

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

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

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

yields
:$\frac = \frac.$

Length of the angle bisector

The length of the angle bisector $d$ can be found by $d^2 = bc - mn = m n (k^2-1) = bc \left( 1-\frac \right)$,

where $k = \frac b n = \frac c m = \frac$ is the constant of proportionality from the angle bisector theorem.

Proof: By Stewart's theorem (which is more general than Apollonius's theorem), we have

$\begin b^2 m + c^2 n &= a(d^2 + mn) \\ (kn)^2 m + (km)^2 n &= a(d^2 + mn) \\ k^2 (m+n)mn &= (m+n) (d^2 + mn) \\ k^2 mn &= d^2 + mn \\ (k^2 - 1) mn &= d^2 \\ \end$

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 intersects the extended side in , the exterior angle bisector in intersects the extended side in and the exterior angle bisector in intersects the extended side in , 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 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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