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.

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 Template:Math. Let the angle bisector of angle Template:Math 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{| B D |}{| C D |} = \frac{| B A |}{| C A |},

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

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

\frac{| B D |}{| C D |} = \frac{| B A | \sin ∠ D A B}{| C A | \sin ∠ D A C} .

This reduces to the previous version if Template:Mvar is the bisector of Template:Math. 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

As shown in the accompanying animation, the theorem can be proved using similar triangles. In the version illustrated here, the triangle $△ A B C$ gets reflected across a line that is perpendicular to the angle bisector $A D$ , resulting in the triangle $△ A B_{2} C_{2}$ with bisector $A D_{2}$ . The fact that the bisection-produced angles $∠ B A D$ and $∠ C A D$ are equal means that $B A C_{2}$ and $C A B_{2}$ are straight lines. This allows the construction of triangle $△ C_{2} B C$ that is similar to $△ A B D$ . Because the ratios between corresponding sides of similar triangles are all equal, it follows that $| A B | / | A C_{2} | = | B D | / | C D |$ . However, $A C_{2}$ was constructed as a reflection of the line $A C$ , and so those two lines are of equal length. Therefore, $| A B | / | A C | = | B D | / | C D |$ , yielding the result stated by the theorem.

Proof using Law of Sines

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

Template:NumBlk

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

\sin ∠ A D B = \sin ∠ A D C .

Angles Template:Math and Template:Math are equal. Therefore, the right hand sides of equations (Template:EquationNote) and (Template:EquationNote) are equal, so their left hand sides must also be equal.

\frac{| B D |}{| C D |} = \frac{| A B |}{| A C |},

which is the angle bisector theorem.

If angles Template:Math are unequal, equations (Template:EquationNote) and (Template:EquationNote) can be re-written as:

\frac{| A B |}{| B D |} \sin ∠ D A B = \sin ∠ A D B,

\frac{| A C |}{| C D |} \sin ∠ D A C = \sin ∠ A D C .

Angles Template:Math are still supplementary, so the right hand sides of these equations are still equal, so we obtain:

\frac{| A B |}{| B D |} \sin ∠ D A B = \frac{| A C |}{| C D |} \sin ∠ D A C,

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

Proof using triangle altitudes

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 Template:Math.

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

Template:Math are right angles, while the angles Template:Math 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 Template:Math are similar (AAA), which implies that:

\frac{| B D |}{| C D |} = \frac{| B B_{1} |}{| C C_{1} |} = \frac{| A B | \sin ∠ B A D}{| A C | \sin ∠ C A D} .

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

\frac{| B D |}{| A B |} = \sin ∠ B A D and \frac{| C D |}{| A C |} = \sin ∠ D A C,

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 Template:Math, which are created by the angle bisector in Template:Mvar. Computing those areas twice using different formulas, that is $\frac{1}{2} g h$ with base $g$ and altitude Template:Mvar and $\frac{1}{2} a b \sin (γ)$ 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 $α$ be half of the angle in Template:Mvar. Then

\frac{| △ A B D |}{| △ A C D |} = \frac{\frac{1}{2} | B D | h}{\frac{1}{2} | C D | h} = \frac{| B D |}{| C D |}

and

\frac{| △ A B D |}{| △ A C D |} = \frac{\frac{1}{2} | A B | | A D | \sin (α)}{\frac{1}{2} | A C | | A D | \sin (α)} = \frac{| A B |}{| A C |}

yields

\frac{| B D |}{| C D |} = \frac{| A B |}{| A C |} .

Length of the angle bisector

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

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

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

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 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:^[1]

\frac{| E B |}{| E C |} = \frac{| A B |}{| A C |}

,

\frac{| F B |}{| F A |} = \frac{| C B |}{| C A |}

,

\frac{| D A |}{| D C |} = \frac{| B A |}{| B C |}

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.^[2]

History

The angle bisector theorem appears as Proposition 3 of Book VI in Euclid's Elements. According to Template:Harvtxt, 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

Template:Expand section 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, Template:ISBN, pp. 3-4
↑ Roger A. Johnson: Advanced Euclidean Geometry. Dover 2007, Template:ISBN, p. 149 (original publication 1929 with Houghton Mifflin Company (Boston) as Modern Geometry).
↑ Template:Cite book
(3 vols.): Template:ISBN (vol. 1), Template:ISBN (vol. 2), Template:ISBN (vol. 3). Heath's authoritative translation plus extensive historical research and detailed commentary throughout the text.

External links

Template:Ancient Greek mathematics

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

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

[3] Template:Cite book
(3 vols.): Template:ISBN (vol. 1), Template:ISBN (vol. 2), Template:ISBN (vol. 3). Heath's authoritative translation plus extensive historical research and detailed commentary throughout the text.

[1]

[2]

[3]

Angle bisector theorem

Contents

Theorem