# Angle bisector theorem En géométrie, 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.

# Angle bisector theorem

Dans géométrie, la 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.

## Théorème[]

Consider a triangle abc. Laisse le angle bisector of angle UN intersect side avant JC at a point entre B et 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 UN B to the length of side CA:

$"{displaystyle$ et conversely, if a point on the side avant JC of triangle abc divides avant JC in the same ratio as the sides UN B et CA, alors UN D is the angle bisector of angle ∠ A.

The generalized angle bisector theorem states that if lies on the line avant JC, alors

$"{displaystyle$sin angle DAB}{|CA|sin angle DAC}}.} This reduces to the previous version if UN D is the bisector of ∠ BAC. Lorsque is external to the segment avant JC, 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.

## Preuves[]

### Preuve 1[]

In the above diagram, use the law of sines on triangles ABD et ACD:

$"{displaystyle$sin angle ADB}{sin angle DAB}}} (1)

$"{displaystyle$sin angle ADC}{sin angle DAC}}} (2)

Since supplementary angles have equal sines,

$"{displaystyle$sin angle ADB}={sin angle ADC}.} Angles ∠ DAB et ∠ DAC are equal. Par conséquent, the right hand sides of equations (1) et (2) are equal, so their left hand sides must also be equal.

$"{displaystyle$ which is the angle bisector theorem.

If angles ∠ DAB et ∠ DAC are unequal, equations (1) et (2) can be re-written as:

$"{displaystyle$sin angle DAB=sin angle ADB},} $"{displaystyle$sin angle DAC=sin angle ADC}.} Angles ∠ ADB et ∠ ADC are still supplementary, so the right hand sides of these equations are still equal, so we obtain:

$"{displaystyle$sin angle DAB={frac {|CA|}{|CD|}}sin angle DAC},} which rearranges to the "generalized" version of the theorem.

### Preuve 2[]

Laisser be a point on the line avant JC, not equal to B ou C et telle que UN D is not an altitude of triangle abc.

Laisser B1 be the base (foot) of the altitude in the triangle ABD à travers B et laissez C1 be the base of the altitude in the triangle ACD à travers C. Alors, si is strictly between B et C, one and only one of B1 ou C1 lies inside triangle abc and it can be assumed sans perte de généralité ce B1 does. This case is depicted in the adjacent diagram. Si lies outside of segment avant JC, then neither B1 nor C1 lies inside the triangle.

∠ DB1B et ∠ DC1C are right angles, while the angles ∠ B1DB et ∠ C1CC are congruent if lies on the segment avant JC (C'est, entre B et C) and they are identical in the other cases being considered, so the triangles DB1B et CC1C are similar (AAA), ce qui implique que:

$"{displaystyle$BB_{1}|}{|CC_{1}|}}={frac {|UN B|sin angle BAD}{|CA|sin angle CAD}}.} Si is the foot of an altitude, alors,

$"{displaystyle$=sin angle BAD{texte{ et }}{frac {|CD|}{|CA|}}=sin angle DAC,} and the generalized form follows.

### Preuve 3[] $"{displaystyle$displaystyle alpha ={tfrac {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

$"{displaystyle$displaystyle triangle BAD} et

$"{displaystyle$displaystyle triangle CAD} , which are created by the angle bisector in

$"{displaystyle$ . Computing those areas twice using different formulas, C'est

$"{displaystyle$gh} with base

$"{displaystyle$ and altitude

$"{displaystyle$ et

$"{displaystyle$absin(gamma )} with sides

$"{displaystyle$ ,

$"{displaystyle$ and their enclosed angle

$"{displaystyle$ , will yield the desired result.

Laisser

$"{displaystyle$ denote the height of the triangles on base

$"{displaystyle$ et

$"{displaystyle$ be half of the angle in

$"{displaystyle$ . Alors

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

$"{displaystyle$triangle ABD|}{|triangle ACD|}}={frac {{frac {1}{2}}|UN B||UN D|péché(alpha )}{{frac {1}{2}}|CA||UN D|péché(alpha )}}={frac {|UN B|}{|CA|}}} donne

$"{displaystyle$ ## Exterior angle bisectors[] exterior angle bisectors (dotted red):
Points , E, F are collinear and the following equations for ratios hold:

$"{displaystyle$EB|}{|CE|}}={tfrac {|UN B|}{|CA|}}} ,

$"{displaystyle$FB|}{|FA|}}={tfrac {|CB|}{|Californie|}}} ,

$"{displaystyle$DA|}{|CC|}}={tfrac {|BA|}{|avant JC|}}} 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

$"{displaystyle$ intersects the extended side

$"{displaystyle$ dans

$"{displaystyle$ , the exterior angle bisector in

$"{displaystyle$ intersects the extended side

$"{displaystyle$ dans

$"{displaystyle$ and the exterior angle bisector in

$"{displaystyle$ intersects the extended side

$"{displaystyle$ dans

$"{displaystyle$ , then the following equations hold:

$"{displaystyle$EB|}{|CE|}}={frac {|UN B|}{|CA|}}} ,

$"{displaystyle$FB|}{|FA|}}={frac {|CB|}{|Californie|}}} ,

$"{displaystyle$DA|}{|CC|}}={frac {|BA|}{|avant JC|}}} The three points of intersection between the exterior angle bisectors and the extended triangle sides

$"{displaystyle$ ,

$"{displaystyle$ et

$"{displaystyle$ sont colinéaires, that is they lie on a common line.

## Histoire[]

The angle bisector theorem appears as Proposition 3 of Book VI in Euclid's Elements. According to Heath (1956, p. 197 (volume. 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:

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; et, 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 avec: more theorems/results. You can help by adding to it. (Septembre 2020)

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

## Références[]

1. ^ Alfred S. Posamentier: Géométrie euclidienne avancée: Excursions for Students and Teachers. Springer, 2002,

ISBN 9781930190856, pp. 3-4

2. ^ Roger A.. Johnson: Géométrie euclidienne avancée. Douvres 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: la presse de l'Universite de Cambridge, 1925] éd.). New York: Publications de Douvres.
(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.

## Liens externes[]

Si vous voulez connaître d'autres articles similaires à Angle bisector theorem vous pouvez visiter la catégorie Elementary geometry.

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