Angle bisector theorem

File:Triangle ABC with bisector AD.svg

The theorem states for any triangle ∠ DAB and ∠ DAC where AD is a bisector, then $|BD|:|CD|=|AB|:|AC|.$

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.

Theorem

Consider a triangle △ABC. Let the angle bisector of angle ∠ A intersect side Template:Mvar at a point Template:Mvar between Template:Mvar and Template:Mvar. The angle bisector theorem states that the ratio of the length of the line segment Template:Mvar to the length of segment Template:Mvar is equal to the ratio of the length of side Template:Mvar to the length of side Template:Mvar:

$\frac{|BD|}{|CD|}=\frac{|AB|}{|AC|},$

and conversely, if a point Template:Mvar on the side Template:Mvar of △ABC divides Template:Mvar in the same ratio as the sides Template:Mvar and Template:Mvar, then Template:Mvar is the angle bisector of angle ∠ A.

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

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

This reduces to the previous version if Template:Mvar is the bisector of ∠ BAC. When Template:Mvar is external to the segment Template:Mvar, 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

File:Animated illustration of angle bisector theorem.gif

Animated illustration of the angle bisector theorem.

As shown in the accompanying animation, the theorem can be proved using similar triangles. In the version illustrated here, the triangle $\mathrm{△}ABC$ gets reflected across a line that is perpendicular to the angle bisector $AD$, resulting in the triangle $\mathrm{△}A{B}_2{C}_2$ with bisector $A{D}_2$. The fact that the bisection-produced angles $\mathrm{\angle }BAD$ and $\mathrm{\angle }CAD$ are equal means that $BA{C}_2$ and $CA{B}_2$ are straight lines. This allows the construction of triangle $\mathrm{△}{C}_{2}BC$ that is similar to $\mathrm{△}ABD$. Because the ratios between corresponding sides of similar triangles are all equal, it follows that $|AB|/|A{C}_2|=|BD|/|CD|$. However, $A{C}_2$ 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 △ABD and △ACD:

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

$\sin \mathrm{\angle }ADB=\sin \mathrm{\angle }ADC.$

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.

$\frac{|BD|}{|CD|}=\frac{|AB|}{|AC|},$

which is the angle bisector theorem.

If angles ∠ DAB, ∠ DAC are unequal, equations (1) and (2) can be re-written as:

$\frac{|AB|}{|BD|}\sin \mathrm{\angle }DAB=\sin \mathrm{\angle }ADB,$

$\frac{|AC|}{|CD|}\sin \mathrm{\angle }DAC=\sin \mathrm{\angle }ADC.$

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

$\frac{|AB|}{|BD|}\sin \mathrm{\angle }DAB=\frac{|AC|}{|CD|}\sin \mathrm{\angle }DAC,$

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

Proof using triangle altitudes

File:Bisekt.svg

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

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

∠ DB₁B, ∠ DC₁C are right angles, while the angles ∠ B₁DB, ∠ C₁DC are congruent if Template:Mvar lies on the segment Template:Mvar (that is, between Template:Mvar and Template:Mvar) and they are identical in the other cases being considered, so the triangles △DB₁B, △DC₁C are similar (AAA), which implies that:

$\frac{|BD|}{|CD|}=\frac{|B{B}_1|}{|C{C}_1|}=\frac{|AB|\sin \mathrm{\angle }BAD}{|AC|\sin \mathrm{\angle }CAD}.$

If Template:Mvar is the foot of an altitude, then,

$\frac{|BD|}{|AB|}=\sin \mathrm{\angle }BAD\text{ and }\frac{|CD|}{|AC|}=\sin \mathrm{\angle }DAC,$

and the generalized form follows.

Proof using triangle areas

File:Angle bisector proof.svg

$\alpha =\frac{\mathrm{\angle }BAC}{2}=\mathrm{\angle }BAD=\mathrm{\angle }CAD$

A quick proof can be obtained by looking at the ratio of the areas of the two triangles △BAD, △CAD, which are created by the angle bisector in Template:Mvar. Computing those areas twice using different formulas, that is $\frac{1}{2}gh$ with base $g$ and altitude Template:Mvar and $\frac{1}{2}ab\sin (\gamma )$ with sides Template:Mvar and their enclosed angle Template:Mvar, will yield the desired result.

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

$\frac{|\mathrm{△}ABD|}{|\mathrm{△}ACD|}=\frac{\frac{1}{2}|BD|h}{\frac{1}{2}|CD|h}=\frac{|BD|}{|CD|}$

and

$\frac{|\mathrm{△}ABD|}{|\mathrm{△}ACD|}=\frac{\frac{1}{2}|AB||AD|\sin (\alpha )}{\frac{1}{2}|AC||AD|\sin (\alpha )}=\frac{|AB|}{|AC|}$

yields

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

Length of the angle bisector

File:Stewarts theorem.svg

Diagram of Stewart's theorem

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

where $k=\frac{b}{n}=\frac{c}{m}=\frac{b+c}{a}$ 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

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

Exterior angle bisectors

File:Aussenwinkelhalbierende2.svg

exterior angle bisectors (dotted red):
Points Template:Mvar are collinear and the following equations for ratios hold:
$\frac{|EB|}{|EC|}=\frac{|AB|}{|AC|}$, $\frac{|FB|}{|FA|}=\frac{|CB|}{|CA|}$, $\frac{|DA|}{|DC|}=\frac{|BA|}{|BC|}$

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 Template:Mvar intersects the extended side Template:Mvar in Template:Mvar, the exterior angle bisector in Template:Mvar intersects the extended side Template:Mvar in Template:Mvar and the exterior angle bisector in Template:Mvar intersects the extended side Template:Mvar in Template:Mvar, then the following equations hold:

$\frac{|EB|}{|EC|}=\frac{|AB|}{|AC|}$, $\frac{|FB|}{|FA|}=\frac{|CB|}{|CA|}$, $\frac{|DA|}{|DC|}=\frac{|BA|}{|BC|}$

The three points of intersection between the exterior angle bisectors and the extended triangle sides Template:Mvar 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 who noted that Pappus assumed this result without proof. Heath goes on to say that Augustus De Morgan proposed that the two statements should be combined as follows:

If an angle of a triangle is bisected internally or externally by a straight line which cuts the opposite side or the opposite side produced, the segments of that side will have the same ratio as the other sides of the triangle; and, if a side of a triangle be divided internally or externally so that its segments have the same ratio as the other sides of the triangle, the straight line drawn from the point of section to the angular point which is opposite to the first mentioned side will bisect the interior or exterior angle at that angular point.

Applications

This theorem has been used to prove the following theorems/results:

• Coordinates of the incenter of a triangle
• Circles of Apollonius

Further reading

• G. W. I. S. Amarasinghe: On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem, Global Journal of Advanced Research on Classical and Modern Geometries, Vol. 01(01), pp. 15–27, 2012.

External links

• A Property of Angle Bisectors at cut-the-knot
• Intro to angle bisector theorem at Khan Academy