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Angle bisector theorem

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Geometrical theorem relating the lengths of two segments that divide a triangle

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

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

Ingeometry, theangle bisector theorem is concerned with the relativelengths of the twosegments that atriangle's side is divided into by aline thatbisects the oppositeangle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

Theorem

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Consider a triangle△ABC. Let theangle bisector of angle∠Aintersect sideBC at a pointD betweenB andC. The angle bisector theorem states that the ratio of the length of theline segmentBD to the length of segmentCD is equal to the ratio of the length of sideAB to the length of sideAC:

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

andconversely, if a pointD on the sideBC of△ABC dividesBC in the same ratio as the sidesAB andAC, thenAD 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 ifD lies on the lineBC, then

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

This reduces to the previous version ifAD is the bisector of∠BAC. WhenD is external to the segmentBC, 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

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

Proof using similar triangles

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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 $A D {\displaystyle AD}$ , 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 $A C {\displaystyle AC}$ , 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

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In the above diagram, use thelaw of sines on triangles△ABD and△ACD:

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

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

Angles∠ADB and∠ADC form a linear pair, that is, they are adjacentsupplementary 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

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LetD be a point on the lineBC, not equal toB orC and such thatAD is not analtitude of triangle△ABC.

LetB₁ be the base (foot) of the altitude in the triangle△ABD throughB and letC₁ be the base of the altitude in the triangle△ACD throughC. Then, ifD is strictly betweenB andC, one and only one ofB₁ orC₁ lies inside△ABC and it can be assumedwithout loss of generality thatB₁ does. This case is depicted in the adjacent diagram. IfD lies outside of segmentBC, then neitherB₁ norC₁ lies inside the triangle.

∠DB₁B, ∠DC₁C are right angles, while the angles∠B₁DB, ∠C₁DC are congruent ifD lies on the segmentBC (that is, betweenB andC) 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 {|BB_{1}|}{|CC_{1}|}}={\frac {|AB|\sin \angle BAD}{|AC|\sin \angle CAD}}.

IfD 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 isosceles triangles

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Construct point $D^{'} {\displaystyle D'}$ on the bisector such that $\triangle ABD\sim \triangle ACD'$ . We aim to show that $|CD|=|CD'|$ .

In the case where $D^{'} {\displaystyle D'}$ lies on ${\overline {AD}}$ , we have that $\angle CD'D=180^{\circ }-\angle CD'A=180^{\circ }-\angle BDA=\angle CDD',$ and in the case where $D^{'} {\displaystyle D'}$ does not lie on ${\overline {AD}}$ , we have that $\angle CD'D=\angle CD'A=\angle BDA=\angle CDD'.$ Either way, $\triangle CDD'$ is isosceles, implying that $|CD|=|CD'|$ . Therefore, ${\frac {|AB|}{|AC|}}={\frac {|BD|}{|CD'|}}={\frac {|BD|}{|CD|}},$ which was the desired result.

Proof using triangle areas

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

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 inA. Computing those areas twice usingdifferent formulas, that is ${\tfrac {1}{2}}gh$ with base $g {\displaystyle g}$ and altitudeh and ${\tfrac {1}{2}}ab\sin(\gamma )$ with sidesa, b and their enclosed angleγ, will yield the desired result.

Leth denote the height of the triangles on baseBC and $\alpha$ be half of the angle inA. 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

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The length of the angle bisector $d {\displaystyle d}$ 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: ByStewart's theorem (which is more general thanApollonius'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

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{\displaystyle {\tfrac {|EB|}{|EC|}}={\tfrac {|AB|}{|AC|}}} — exterior angle bisectors (dotted red):
PointsD, E, F are collinear and the following equations for ratios hold:
${\tfrac {|EB|}{|EC|}}={\tfrac {|AB|}{|AC|}}$ , ${\tfrac {|FB|}{|FA|}}={\tfrac {|CB|}{|CA|}}$ , ${\tfrac {|DA|}{|DC|}}={\tfrac {|BA|}{|BC|}}$

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 inA intersects the extended sideBC inE, the exterior angle bisector inB intersects the extended sideAC inD and the exterior angle bisector inC intersects the extended sideAB inF, 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 sidesD, E, F are collinear, that is they lie on a common line.^[2]

History

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The angle bisector theorem appears as Proposition 3 of Book VI inEuclid's Elements. According toHeath (1956, p. 197 (vol. 2)), the corresponding statement for an external angle bisector was given byRobert Simson who noted thatPappus assumed this result without proof. Heath goes on to say thatAugustus 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

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This sectionneeds expansion with: more theorems/results. You can help byadding missing information.(September 2020)

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

Coordinates of theincenter of a triangle
Circles of Apollonius

References

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^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) asModern 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

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InElements	Angle bisector theorem Exterior angle theorem Euclidean algorithm Euclid's theorem Geometric mean theorem Hinge theorem Inscribed angle theorem Intercept theorem Intersecting chords theorem Intersecting secants theorem Law of cosines Pons asinorum Pythagorean theorem Tangent-secant theorem Thales's theorem Theorem of the gnomon

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