Difference between revisions of "Carnot's Theorem"
(here's a proof!)
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Revision as of 08:51, 3 July 2011
Only if: Assume that the given perpendiculars are concurrent at . Then, from the Pythagorean Theorem, , , , , , and . Substituting each and every one of these in and simplifying gives the desired result.
If: Consider the intersection of the perpendiculars from and . Call this intersection point , and let be the perpendicular from to . From the other direction of the desired result, we have that . We also have that , which implies that . This is a difference of squares, which we can easily factor into . Note that , so we have that . This implies that , which gives the desired result.
is a triangle. Take points on the perpendicular bisectors of respectively. Show that the lines through perpendicular to respectively are concurrent. (Source)