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

As shown in the accompanying animation, the theorem can be proved using similar triangles. In the version illustrated here, the triangle ${\displaystyle \triangle ABC}$ gets reflected across a line that is perpendicular to the angle bisector ${\displaystyle AD}$, resulting in the triangle ${\displaystyle \triangle AB_{2}C_{2}}$ with bisector ${\displaystyle AD_{2}}$. The fact that the bisection-produced angles ${\displaystyle \angle BAD}$ and ${\displaystyle \angle CAD}$ are equal means that ${\displaystyle BAC_{2}}$ and ${\displaystyle CAB_{2}}$ are straight lines. This allows the construction of triangle ${\displaystyle \triangle C_{2}BC}$ that is similar to ${\displaystyle \triangle ABD}$. Because the ratios between corresponding sides of similar triangles are all equal, it follows that ${\displaystyle |AB|/|AC_{2}|=|BD|/|CD|}$. However, ${\displaystyle AC_{2}}$ was constructed as a reflection of the line ${\displaystyle AC}$, and so those two lines are of equal length. Therefore, ${\displaystyle |AB|/|AC|=|BD|/|CD|}$, 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:

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

(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

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

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, ∠ 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, ∠ 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 using triangle altitudes

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 △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, ∠ DC₁C are right angles, while the angles ∠ B₁DB, ∠ 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, △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 using triangle areas

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 A. Computing those areas twice using different formulas, that is ${\displaystyle {\tfrac {1}{2}}gh}$ with base g and altitude h and ${\displaystyle {\tfrac {1}{2}}ab\sin(\gamma )}$ with sides a, b and their enclosed angle γ, will yield the desired result.

Let h denote the height of the triangles on base BC and ${\displaystyle \alpha }$ be half of the angle in A. Then

${\displaystyle {\frac {|\triangle ABD|}{|\triangle ACD|}}={\frac {{\frac {1}{2}}|BD|h}{{\frac {1}{2}}|CD|h}}={\frac {|BD|}{|CD|}}}$

and

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

yields

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

Length of the angle bisector

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

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

Proof: By Stewart's theorem, we have

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

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:^[1]

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

The three points of intersection between the exterior angle bisectors and the extended triangle sides D, E, F are collinear, that is they lie on a common line.^[2]

History

The angle bisector theorem appears as Proposition 3 of Book VI in Euclid's Elements. According to Heath (1956, p. 197 (vol. 2)), 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:^[3]

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 section needs expansion with: more theorems/results. You can help by adding to it. (September 2020)

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

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

References