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Angle bisector

This is an AoPSWiki Word of the Week for June 6-12

Angle Bisector

For an angle $\angle BAC$, the (internal) angle bisector of $\angle BAC$ is the line from $A$ such that the angle between this line and $\overline{AB}$ is congruent to the angle between this line and $\overline{AC}$:

Anglebisector.png

An angle $\angle BAC$ also has an external angle bisector, which bisects external angle $BAC$:

[asy] import markers; pair A,B,C,D,E,F; B=(0,0); C=(5,0); A=(4,2); D=(3,4); E=(10/3,0); F=rotate(-90,A)*(E); draw(A--B--C--cycle,blue); draw(C--D,blue); pen p=blue+0.15mm; draw(A--F,p); dot(A^^B^^C,red); label("$A$",A,NE); label("$B$",B,W); label("$C$",C,E); markangle(n=1,radius=20,D,A,F,green); markangle(n=1,radius=22,F,A,B,green); [/asy]

The external angle is defined by \[\angle A + \text{ external } \angle A = 180^\circ,\] and the two angle bisectors are perpendicular to each other.

Features of Angle Bisectors

  • The distances from a point on an angle bisector to both of its sides are equal.
  • The angle bisectors are the locus of points which are equidistant from the two sides of the angle.
  • A reflection about either angle bisector maps the two sides of the angle to each other.
  • In a triangle, the Angle Bisector Theorem gives the ratio in which the angle bisector cuts the opposite side.
  • In a triangle, the internal angle bisectors (which are cevians) all intersect at the incenter of the triangle. The internal angle bisector of one angle and the external angle bisectors of the other two angles all intersect at an excenter of the triangle.
  • A bisector of an angle can be constructed using a compass and straightedge.

[asy] size(300); import markers; pair excenter(pair A, pair B, pair C){ pair X, Z; X=A+expi((angle(A-B)+angle(C-A))/2); Z=C+expi((angle(C-B)+angle(A-C))/2); return extension(X,A,Z,C); } pair X=(0,0), Y=(-10,0), Z=(-3,6); pair exX=excenter(Z,X,Y), exY=excenter(X,Y,Z), exZ=excenter(Y,Z,X); label("X",X,2*(1.5,-1));label("Y",Y,2*(-3,-1));label("Z",Z,(0.6,2)); dot(X^^Y^^Z); draw(0.3*(X-Y)+X--Y+0.3*(Y-X));draw(0.3*(Y-Z)+Y--Z+0.3*(Z-Y));draw(0.3*(X-Z)+X--Z+0.3*(Z-X)); draw(X+0.3*(X-exX)--exX,orange);draw(Y+0.3*(Y-exY)--exY,orange);draw(Z+0.3*(Z-exZ)--0.6*(exZ-Z)+Z,orange); draw(Y-0.5*(exX-Y)--Y+0.5*(exX-Y),green);draw(Z-0.5*(exX-Z)--Z+0.5*(exX-Z),green);draw(X-0.5*(exY-X)--X+0.5*(exY-X),green); pair I=extension(X,exX,Z,exZ); dot("I",I,(-1,2)); draw(circle(I,length(I-foot(I,X,Y))),blue); markangle(n=1,3*(Z-X)+X,Z,exX,radius=22,marker(markinterval(stickframe(n=3),true))); markangle(n=1,exX,Z,Y,radius=20,marker(markinterval(stickframe(n=3),true))); markangle(n=1,3*(Y-Z)+Z,Y,exZ,radius=22,marker(markinterval(stickframe(n=2),true))); markangle(n=1,exZ,Y,X,radius=20,marker(markinterval(stickframe(n=2),true))); markangle(n=1,3*(X-Y)+Y,X,exY,radius=22,marker(markinterval(stickframe(n=1),true))); markangle(n=1,exY,X,Z,radius=20,marker(markinterval(stickframe(n=1),true))); markangle(n=3,I,Z,X,radius=20); markangle(n=3,Y,Z,I,radius=22); markangle(n=2,Z,X,I,radius=20); markangle(n=2,I,X,Y,radius=22); markangle(n=1,X,Y,I,radius=22); markangle(n=1,I,Y,Z,radius=20); [/asy]

Enlarge.png
Triangle $\triangle XYZ$ with incenter I, incircle (blue), angle bisectors (orange), and external angle bisectors (green)

Proofs

Angle Bisector Locus Theorem

Theorem: A point lies on the internal angle bisector of $\angle BAC$ if and only if it is equidistant from the sides $\overline{AB}$ and $\overline{AC}$.

Proof:

Let point $D$ lie on the angle bisector of $\angle BAC$. Drop perpendiculars from $D$ to lines $AB$ and $AC$, meeting them at points $E$ and $F$ respectively.

Since \[\angle DAE = \angle DAC,\] and both triangles $\triangle DAE$ and $\triangle DAF$ are right triangles with a shared hypotenuse $AD$ and equal angles, they are congruent by AAS. Hence, \[DE = DF.\]

Conversely, if $D$ is equidistant from $AB$ and $AC$, then $\triangle DAE \cong \triangle DAF$, and $D$ lies on the angle bisector of $\angle BAC$.

Angle Bisector Theorem

Theorem: In triangle $\triangle ABC$, if the internal angle bisector of $\angle A$ intersects $\overline{BC}$ at point $D$, then \[\frac{BD}{DC} = \frac{AB}{AC}.\]

Proof:

Let $\angle BAD = \angle DAC$, and draw the angle bisector $AD$.

Construct a line through $C$ parallel to $AD$, and let it intersect the extension of $AB$ at point $E$.

Since \[\angle BAD = \angle DAC = \angle ECA\] (by alternate interior angles), triangles $\triangle ABD$ and $\triangle ECD$ are similar by AA.

Therefore, \[\frac{BD}{DC} = \frac{AB}{AC}.\]

Internal and External Angle Bisectors Are Perpendicular

Theorem: The internal and external angle bisectors of any angle are perpendicular.

Proof:

Let the internal and external angle bisectors divide $\angle A$ and its supplement into two equal parts.

The internal angle is $\theta$, and the external angle is $180^\circ - \theta$.

So the internal bisector makes an angle of $\theta/2$ with one side, and the external bisector makes an angle of $(180^\circ - \theta)/2$ with the same side. Adding them: \[\frac{\theta}{2} + \frac{180^\circ - \theta}{2} = \frac{180^\circ}{2} = 90^\circ,\] so the two bisectors are perpendicular.

See also

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