11159
EXPLORE
LATEST
ABOUT
AUTHORING AREA
PARTICIPATE
Your browser does not support JavaScript or it may be disabled!
Division of an Angle Bisector by the Incenter
Let ABC be a triangle with incenter I. Let AA' be the bisector of
with A' on BC. Then
.
Contributed by:
Jay Warendorff
THINGS TO TRY
Drag Locators
SNAPSHOTS
DETAILS
Problem 1.17(b) in V. Prasolov,
Problems in Plane and Solid Geometry
, Vol. 1,
Plane Geometry
[PDF], (D. Leites, ed. and trans.).
RELATED LINKS
Angle Bisector
(
Wolfram
MathWorld
)
Incenter
(
Wolfram
MathWorld
)
Incircle
(
Wolfram
MathWorld
)
PERMANENT CITATION
"
Division of an Angle Bisector by the Incenter
" from
the Wolfram Demonstrations Project
http://demonstrations.wolfram.com/DivisionOfAnAngleBisectorByTheIncenter/
Contributed by:
Jay Warendorff
Share:
Embed Interactive Demonstration
New!
Just copy and paste this snippet of JavaScript code into your website or blog to put the live Demonstration on your site.
More details »
Download Demonstration as CDF »
Download Author Code »
(preview »)
Files require
Wolfram
CDF Player
or
Mathematica
.
Related Demonstrations
More by Author
Division of the Opposite Side by an Angle Bisector
Jay Warendorff
Angle Bisector Theorem
Jay Warendorff
The Intersection of an Angle Bisector and a Perpendicular Bisector
Jay Warendorff
Concyclic Points Associated with an Angle Bisector and an Excircle
Jay Warendorff
Angle Bisectors in a Triangle
Jay Warendorff
Two Triangles of Equal Area on Either Side of an Angle Bisector
Jay Warendorff
A Concurrency Generated by the Angle Bisectors
Jay Warendorff
Perpendiculars from a Point on the Line between the Endpoints of the Angle Bisectors
Jay Warendorff
Bisectors of the Angles of the Orthic Triangle
Jay Warendorff
Rhombi at the Incenter of a Triangle
Jay Warendorff
Related Topics
Plane Geometry
Triangles
Browse all topics
Note: To run this Demonstration you need Mathematica 7+ or the free Mathematica Player 7EX
Download or upgrade to
Mathematica Player 7EX
I already have
Mathematica Player
or
Mathematica 7+