11266
EXPLORE
LATEST
ABOUT
AUTHORING AREA
PARTICIPATE
Your browser does not support JavaScript or it may be disabled!
A Lemma of Archimedes about a Bisected Segment
Let AB be the diameter of a semicircle. Let CB and CD be tangents. If DE is perpendicular to AB and DE intersects AC at F then DF = FE.
Drag the red point to change the figure.
Contributed by:
Jay Warendorff
After work by:
Antonio Gutierrez
THINGS TO TRY
Drag Locators
SNAPSHOTS
DETAILS
The statement of the theorem is in
Archimedes' Book of Lemmas: Proposition 2
.
RELATED LINKS
Circle Tangent Line
(
Wolfram
MathWorld
)
Perpendicular
(
Wolfram
MathWorld
)
PERMANENT CITATION
"
A Lemma of Archimedes about a Bisected Segment
" from
the Wolfram Demonstrations Project
http://demonstrations.wolfram.com/ALemmaOfArchimedesAboutABisectedSegment/
Contributed by:
Jay Warendorff
After work by:
Antonio Gutierrez
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
The See-Saw Lemma
Jay Warendorff
Tangent Points on a Semicircle
Jay Warendorff
Tangent Circles and Parallel Diameters
Jay Warendorff
Perpendiculars to a Chord
Jay Warendorff
Inscribed and Central Angles in a Circle
Jay Warendorff
Equality of a Segment and an Arc in Archimedes's Spiral
Izidor Hafner
Archimedes' Approximation of Pi
John Tucker
Archimedes's Neusis Angle-Trisection
Izidor Hafner
The Trammel of Archimedes
Jamie Wickham-Jones
Salinon
Michael Schreiber
Related Topics
Greek Mathematics
Plane Geometry
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+