Angle bisector theorem

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.

Note that this theorem is not to be confused with the Inscribed Angle Theorem, which also involves angle bisection (but of an angle of a triangle inscribed in a circle).

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 $AC$ :

{\frac {|BD|}{|CD|}}={\frac {|BA|}{|CA|}},

and conversely, if a point $D$ on the side $BC$ of $△ ABC$ 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

{\frac {|BD|}{|CD|}}={\frac {|BA|\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

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

Proof using similar triangles

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 $\triangle ABC$ gets reflected across a line that is perpendicular to the angle bisector $AD$ , resulting in the triangle $\triangle AB_{2}C_{2}$ with bisector $AD_{2}$ . The fact that the bisection-produced angles $\angle BAD$ and $\angle CAD$ are equal means that $BAC_{2}$ and $CAB_{2}$ are straight lines. This allows the construction of triangle $\triangle C_{2}BC$ 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 $△ ABD$ and $△ ACD$ :

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

(1)

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

(2)

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

{\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.

{\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 \angle DAB=\sin \angle ADB},

{{\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:

{{\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 $B 1$ 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 $△ ABC$ 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.

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

{\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,

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

and the generalized form follows.

Proof using triangle areas

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

A quick proof can be obtained by looking at the ratio of the areas of the two triangles $△ BAD, △ CAD$ , which are created by the angle bisector in $A$ . Computing those areas twice using different formulas, that is ${\tfrac {1}{2}}gh$ with base $g$ and altitude $h$ and ${\tfrac {1}{2}}ab\sin(\gamma )$ 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 $\alpha$ be half of the angle in $A$ . Then

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

and

{\frac {|\triangle ABD|}{|\triangle 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

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

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, we have

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

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]

{\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 $D, E, F$ 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 theorem has been used to prove the following theorems/results:

Coordinates of the incenter of a triangle
Circles of Apollonius

References

^ Alfred S. Posamentier: Advanced Euclidean Geometry: Excursions for Students and Teachers. Springer, 2002, ISBN 9781930190856, pp. 3-4
^ Roger A. Johnson: Advanced Euclidean Geometry. Dover 2007, ISBN 978-0-486-46237-0, p. 149 (original publication 1929 with Houghton Mifflin Company (Boston) as Modern Geometry).
^ Heath, Thomas L. (1956). The Thirteen Books of Euclid's Elements (2nd ed. [Facsimile. Original publication: Cambridge University Press, 1925] ed.). New York: Dover Publications.
(3 vols.): ISBN 0-486-60088-2 (vol. 1), ISBN 0-486-60089-0 (vol. 2), ISBN 0-486-60090-4 (vol. 3). Heath's authoritative translation plus extensive historical research and detailed commentary throughout the text.

External links

[1] Alfred S. Posamentier: Advanced Euclidean Geometry: Excursions for Students and Teachers. Springer, 2002, ISBN 9781930190856, pp. 3-4

[2] Roger A. Johnson: Advanced Euclidean Geometry. Dover 2007, ISBN 978-0-486-46237-0, p. 149 (original publication 1929 with Houghton Mifflin Company (Boston) as Modern Geometry).

[3] Heath, Thomas L. (1956). The Thirteen Books of Euclid's Elements (2nd ed. [Facsimile. Original publication: Cambridge University Press, 1925] ed.). New York: Dover Publications.
(3 vols.): ISBN 0-486-60088-2 (vol. 1), ISBN 0-486-60089-0 (vol. 2), ISBN 0-486-60090-4 (vol. 3). Heath's authoritative translation plus extensive historical research and detailed commentary throughout the text.

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