Angle Bisector

Author

Eric W. Weisstein
November 10, 2021

This notebook downloaded from http://mathworld.wolfram.com/notebooks/PlaneGeometry/AngleBisector.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/AngleBisector.html.

©2021 Wolfram Research, Inc. except for portions noted otherwise

Figure

V5

v={#/Norm[#]&@{.4,.7},{0,0},{1,0}};Show[Graphics[{
{Blue,Line[v]},
{Red,AngleBisectors[Triangle[v]][[2]]},
{Red,EqualAngleDivisionTicks[v,2,1,.2,.02,.1]}
}],AspectRatio->Automatic,PlotRange->All]

V6

AngleBisectorGraphic[q_,opts___?OptionQ]:=With[{x=.2,dx=.02},Graphics[{
Blue,Line[{{Cos[q],Sin[q]},{0,0},{1,0}}],
Red,Circle[{0,0},.2,{0,q}],Line[{{0,0},{Cos[q/2],Sin[q/2]}}],
Table[Line[#{Cos[i q/4],Sin[i q/4]}&/@{x-dx,x+dx}],{i,{1,3}}]
},opts]]

AngleBisectorGraphic[60Degree]

Intersection with Sides (Bisector Triangle)

Intersections@@@Map[TrilinearLine,Transpose[{AngleBisectors[],Sides[ReferenceTriangle]}],{2}]

PerspectiveQ[Triangle[ReferenceTriangle],Triangle[{{0,1,1},{1,0,1},{1,1,0}}]]

Perspector[Triangle[ReferenceTriangle],Triangle[{{0,1,1},{1,0,1},{1,1,0}}]]

Trilinears[Incenter]

Relation to Incircle

t=Triangle[{{0,0},{3,0},{1.3,1.7}}];
Show[Graphics[{
{Red,Incircle[t]},
t,
{Blue,AngleBisectors[t],
EqualAngleDivisionTicks[t,2,.3,.05,.1],
VertexArcs[t,.3]
},
PointLabels[
Join[
Transpose[{VertexPoints[t],{"(Subscript[A, 1])","(Subscript[A, 2])","(Subscript[A, 3])"}}],
{{Incenter[t],"I",Red}},
Append[#,Blue]&/@Transpose[{EndPoint/@AngleBisectors[t],{"(Subscript[T, 1])","(Subscript[T, 2])","(Subscript[T, 3])"}}]
],
PointSize[.025]]},
AspectRatio->Automatic,PlotRange->All]
]

Triangle

Solve[{
(3^2)==x^2+h^2-2x h Cos[α],
5^2==(x+4)^2+h^2-2(x+4)h Cos[α],
8^2==x^2+(x+4)^2-2x(x+4)Cos[2α],
0<α<π},{x,h,α},PositiveReals]

Solve[5/3==(x+4)/x,x]

Manipulate[With[{x1=(-a^2+b^2+c^2)/(2 c),y1=Sqrt[(-a+b+c) (a+b-c) (a-b+c) (a+b+c)]/(2 c)},
Graphics[{EdgeForm[Black],FaceForm[LightYellow],Triangle[{{c,0},{x1,y1},{0,0}}]}]
],{{a,8},.1,10},{{b,6},.1,10},{{c,10},.1,10}]