In this diagram, BD:DC = AB:corrente alternata.

In geometria, il angle bisector theorem is concerned with the relative lengths of the two segments that a triangolo'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.

Teorema[]

Consider a triangle ABC. Lascia il angle bisector di angolo UN intersect side AVANTI CRISTO at a point D fra B e 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 corrente alternata:

CD|}}={frac {|AB|}{|corrente alternata|}},}

e conversely, if a point D on the side AVANTI CRISTO of triangle ABC divides AVANTI CRISTO in the same ratio as the sides AB e corrente alternata, poi ANNO DOMINI is the angle bisector of angle ∠ A.

The generalized angle bisector theorem states that if D lies on the line AVANTI CRISTO, poi

CD|}}={frac {|AB|sin angle DAB}{|corrente alternata|sin angle DAC}}.}

This reduces to the previous version if ANNO DOMINI is the bisector of ∠ BAC. quando D is external to the segment AVANTI CRISTO, 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.

Prove[]

Prova 1[]

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

sin angle ADB}{sin angle DAB}}}

(1)

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

(2)

Angles ∠ ADB e ∠ ADC form a linear pair, questo è, they are adjacent supplementary angles.

Since supplementary angles have equal sines,

sin angle ADB}={sin angle ADC}.}

Angles ∠ DAB e ∠ DAC are equal. Perciò, the right hand sides of equations (1) e (2) are equal, so their left hand sides must also be equal.

CD|}}={frac {|AB|}{|corrente alternata|}},}

which is the angle bisector theorem.

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

sin angle DAB=sin angle ADB},}

CD|}}sin angle DAC=sin angle ADC}.}

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

sin angle DAB={frac {|corrente alternata|}{|CD|}}sin angle DAC},}

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

Prova 2[]

Permettere D be a point on the line AVANTI CRISTO, not equal to B o C e tale che ANNO DOMINI is not an altitude of triangle ABC.

Permettere B₁ be the base (foot) of the altitude in the triangle ABD attraverso B e lascia C₁ be the base of the altitude in the triangle ACD attraverso C. Quindi, Se D is strictly between B e C, one and only one of B₁ o C₁ lies inside triangle ABC and it can be assumed senza perdita di generalità Quello B₁ does. This case is depicted in the adjacent diagram. Se D lies outside of segment AVANTI CRISTO, then neither B₁ nor C₁ lies inside the triangle.

∠ DB₁B e ∠ DC₁C are right angles, while the angles ∠ B₁DB e ∠ C₁DC are congruent if D lies on the segment AVANTI CRISTO (questo è, fra B e C) and they are identical in the other cases being considered, so the triangles DB₁B e DC₁C are similar (AAA), il che lo implica:

CD|}}={frac {|BB_{1}|}{|CC_{1}|}}={frac {|AB|sin angle BAD}{|corrente alternata|sin angle CAD}}.}

Se D is the foot of an altitude, poi,

=sin angle BAD{testo{ e }}{frac {|CD|}{|corrente alternata|}}=sin angle DAC,}

and the generalized form follows.

Prova 3[]

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

e

, which are created by the angle bisector in

. Computing those areas twice using different formulas, questo è

with base

and altitude

e

with sides

,

and their enclosed angle

, will yield the desired result.

Permettere

denote the height of the triangles on base

e

be half of the angle in

. Quindi

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

e

triangle ABD|}{|triangle ACD|}}={frac {{frac {1}{2}}|AB||ANNO DOMINI|peccato(alfa )}{{frac {1}{2}}|corrente alternata||ANNO DOMINI|peccato(alfa )}}={frac {|AB|}{|corrente alternata|}}}

yields

CD|}}={frac {|AB|}{|corrente alternata|}}.}

Exterior angle bisectors[]

exterior angle bisectors (dotted red):
Points D, e, F are collinear and the following equations for ratios hold:

EB|}{|EC|}}={tfrac {|AB|}{|corrente alternata|}}}

,

FB|}{|FA|}}={tfrac {|CB|}{|circa|}}}

,

DA|}{|DC|}}={tfrac {|BA|}{|AVANTI CRISTO|}}}

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

intersects the extended side

in

, the exterior angle bisector in

intersects the extended side

in

and the exterior angle bisector in

intersects the extended side

in

, then the following equations hold:^[1]

EB|}{|EC|}}={frac {|AB|}{|corrente alternata|}}}

,

FB|}{|FA|}}={frac {|CB|}{|circa|}}}

,

DA|}{|DC|}}={frac {|BA|}{|AVANTI CRISTO|}}}

The three points of intersection between the exterior angle bisectors and the extended triangle sides

,

e

are collinear, that is they lie on a common line.^[2]

Storia[]

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; e, 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.

Applicazioni[]

This section needs expansion insieme a: more theorems/results. You can help by adding to it. (settembre 2020)

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

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

Riferimenti[]

Ulteriori letture[]

• G.W.I.S Amarasinghe: 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

link esterno[]

• A Property of Angle Bisectors a cut-the-knot
• Intro to angle bisector theorem a Khan Academy

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