The Semiperimeter and Tangents to Excircles
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Let ABC be a triangle. Construct the excircle opposite A and let D and E be the intersections of the extensions of AB and AC with the excircle. Let be the semiperimeter of ABC. Then = AD = AE.
Contributed by: Jay Warendorff (March 2011)
After work by: Antonio Gutierrez
Open content licensed under CC BY-NC-SA
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Details
The statement of the theorem is in Problem 140. Triangle, Excircle, Tangent, Semiperimeter.
Permanent Citation
"The Semiperimeter and Tangents to Excircles"
http://demonstrations.wolfram.com/TheSemiperimeterAndTangentsToExcircles/
Wolfram Demonstrations Project
Published: March 7 2011