Difference between revisions of "Carnot's Theorem"
|Line 37:||Line 37:|
where r is the [[inradius]] and R is the [[circumradius]]. The sign of the distance is chosen to be negative iff the entire segment OO_i lies outside the triangle.
where r is the [[inradius]] and R is the [[circumradius]]. The sign of the distance is chosen to be negative iff the entire segment OO_ilies outside the triangle.
Revision as of 15:20, 31 May 2018
where is the area of triangle .
Weisstein, Eric W. "Carnot's Theorem." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html
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.
Let be points in the plane of triangle . Then the perpendiculars from to respectively are concurrent if and only if
Let be the feet of perpendiculars from to respectively. Note that from the application of pythogorean theorem to triangles . Now with similar relations for and , Carnot's theorem finishes the job!
is a triangle. Take points on the perpendicular bisectors of respectively. Show that the lines through perpendicular to respectively are concurrent. (Source)