Angle bisector theorem

Consider a triangle {{math|△ABC}}. Let the angle bisector of angle {{math|∠ A}} intersect side {{mvar|{{overline|BC}}}} at a point {{mvar|D}} between {{mvar|B}} and {{mvar|C}}. The angle bisector theorem states that the ratio of the length of the line segment {{mvar|{{overline|BD}}}} to the length of segment {{mvar|{{overline|CD}}}} is equal to the ratio of the length of side {{mvar|{{overline|AB}}}} to the length of side {{mvar|{{overline|AC}}}}:

:$\{\backslash frac$

and conversely, if a point {{mvar|D}} on the side {{mvar|{{overline|BC}}}} of {{math|△ABC}} divides {{mvar|{{overline|BC}}}} in the same ratio as the sides {{mvar|{{overline|AB}}}} and {{mvar|{{overline|AC}}}}, then {{mvar|{{overline|AD}}}} is the angle bisector of angle {{math|∠ A}}.

The generalized angle bisector theorem (which is not necessarily an angle bisector theorem, since the angle {{math|∠ A}} is not necessarily bisected into equal parts) states that if {{mvar|D}} lies on the line {{mvar|{{overline|BC}}}}, then

:$\{\backslash frac$

This reduces to the previous version if {{mvar|{{overline|AD}}}} is the bisector of {{math|∠ BAC}}. When {{mvar|D}} is external to the segment {{mvar|{{overline|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.

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

File:Animated illustration of angle bisector theorem.gif

As shown in the accompanying animation, the theorem can be proved using similar triangles. In the version illustrated here, the triangle $\backslash triangle\; ABC$ gets reflected across a line that is perpendicular to the angle bisector $AD$, resulting in the triangle $\backslash triangle\; A\; B\_2\; C\_2$ with bisector $AD\_2$. The fact that the bisection-produced angles $\backslash angle\; BAD$ and $\backslash angle\; CAD$ are equal means that $BA\; C\_2$ and $CA\; B\_2$ are straight lines. This allows the construction of triangle $\backslash triangle\; C\_2BC$ that is similar to $\backslash triangle\; ABD$. 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.

In the above diagram, use the law of sines on triangles {{math|△ABD}} and {{math|△ACD}}:

{{NumBlk|:|$\{\backslash frac$

{{NumBlk|:|$\{\backslash frac$

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

:$\{\{\backslash sin\; \backslash angle\; ADB\}\}\; =\; \{\backslash sin\; \backslash angle\; ADC\}.$

Angles {{math|∠ DAB}} and {{math|∠ DAC}} are equal. Therefore, the right hand sides of equations ({{EquationNote|1}}) and ({{EquationNote|2}}) are equal, so their left hand sides must also be equal.

:$\{\backslash frac$

which is the angle bisector theorem.

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

:$\{\backslash frac$

:$\{\backslash frac$

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

:$\{\backslash frac$

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

Image:Bisekt.svg

Let {{mvar|D}} be a point on the line {{mvar|BC}}, not equal to {{mvar|B}} or {{mvar|C}} and such that {{mvar|{{overline|AD}}}} is not an altitude of triangle {{math|△ABC}}.

Let {{math|B₁}} be the base (foot) of the altitude in the triangle {{math|△ABD}} through {{mvar|B}} and let {{math|C₁}} be the base of the altitude in the triangle {{math|△ACD}} through {{mvar|C}}. Then, if {{mvar|D}} is strictly between {{mvar|B}} and {{mvar|C}}, one and only one of {{math|B₁}} or {{math|C₁}} lies inside {{math|△ABC}} and it can be assumed without loss of generality that {{math|B₁}} does. This case is depicted in the adjacent diagram. If {{mvar|D}} lies outside of segment {{mvar|{{overline|BC}}}}, then neither {{math|B₁}} nor {{math|C₁}} lies inside the triangle.

{{math|∠ DB₁B, ∠ DC₁C}} are right angles, while the angles {{math|∠ B₁DB, ∠ C₁DC}} are congruent if {{mvar|D}} lies on the segment {{mvar|{{overline|BC}}}} (that is, between {{mvar|B}} and {{mvar|C}}) and they are identical in the other cases being considered, so the triangles {{math|△DB₁B, △DC₁C}} are similar (AAA), which implies that:

:$\{\backslash frac$

and the generalized form follows.

File:Angle bisector proof.svg

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

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

:$$

\frac

\frac{\frac{1}{2}|BD|h}{\frac{1}{2}|CD|h} = \frac

and

:$$

\frac

\frac{\frac{1}{2}|AB||AD|\sin(\alpha)}{\frac{1}{2}|AC||AD|\sin(\alpha)} =

\frac

yields

:$$

\frac

File:Stewarts theorem.svg

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

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

$\backslash begin\{align\}$

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{align}

File:Aussenwinkelhalbierende2.svg

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 {{mvar|A}} intersects the extended side {{mvar|BC}} in {{mvar|E}}, the exterior angle bisector in {{mvar|B}} intersects the extended side {{mvar|AC}} in {{mvar|D}} and the exterior angle bisector in {{mvar|C}} intersects the extended side {{mvar|AB}} in {{mvar|F}}, then the following equations hold:Alfred S. Posamentier: Advanced Euclidean Geometry: Excursions for Students and Teachers. Springer, 2002, {{ISBN|9781930190856}}, pp. [https://books.google.com/books?id=9grsxFZUci8C&pg=PA4 3–4].

:$\backslash frac$

The three points of intersection between the exterior angle bisectors and the extended triangle sides {{mvar|D, E, F}} are collinear, that is they lie on a common line.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).

The angle bisector theorem appears as Proposition 3 of Book VI in Euclid's Elements. According to {{harvtxt|Heath|1956|page=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: {{cite book

| last = Heath

| first = Thomas L.

| author-link = T. L. Heath

| title = The Thirteen Books of Euclid's Elements

| url = https://archive.org/details/thirteenbooksofe00eucl

| url-access = registration

| edition = 2nd ed. [Facsimile. Original publication: Cambridge University Press, 1925]

| year = 1956

| publisher = Dover Publications

| location = New York

}}

: (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.

: 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.

{{Expand section|more theorems/results|date=September 2020}}

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

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

• G. W. I. S. Amarasinghe: [http://www.geometry-math-journal.ro/pdf/Volume1-Issue1/ON%20THE%20STANDARD%20LENGTHS%20OF%20ANGLE%20BISECTORS%20AND%20THE%20ANGLE%20BISECTOR%20THEOREM.pdf 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.

• [https://www.cut-the-knot.org/Curriculum/Geometry/AngleBisectorRatio.shtml A Property of Angle Bisectors] at cut-the-knot
• [http://www.khanacademy.org/math/geometry/hs-geo-similarity/hs-geo-angle-bisector-theorem/v/angle-bisector-theorem-proof Intro to angle bisector theorem] at Khan Academy

{{Ancient Greek mathematics}}

Category:Articles containing proofs

Category:Elementary geometry

Category:Theorems about triangles