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Carnot's Theorem

Let points A', B', and C' be located on the sides BC, AC, and respectively AB of ABC. The perpendiculars to the sides of the triangle at points A', B', and C' are concurrent iff

AC'² - BC'² + BA'² - CA'² + CB'² - AB'² = 0.


First we'll prove the "only if" direction.  Assume the three perpendiculars meet in point P.  The lines AP, BP, CP, A'P, B'P, C'P split the triangle into six right triangles, three pairs of which share a hypotenuse, while three other pairs share a leg.  By the Pythagorean theorem,

AC'² + C'P² = AP²
-BC'² - C'P² = -BP²
BA'² + A'P² = BP²
-CA'² - A'P² = -CP²
CB'² + B'P² = CP²
-AB'² - B'P² = -AP²

Adding up the six equations completes the proof.

Now, to tackle the "if" direction, assume

AC'² - BC'² + BA'² - CA'² + CB'² - AB'² = 0.

and let P be the point of intersection of A'P and B'P.  Drop a perpendicular DP from P to AB.  By the "only if" part, already proven,

AD² - BD² + BA'² - CA'² + CB'² - AB'² = 0.

and then by subtracting the last two equations, one from the other, we get

AD² - BD² = AC'² - BC'² 
(AD -BD)(AD+BD) = (AC' -BC')(AC'+BC') 
(AD -BD)(AB) = (AC' -BC')(AB) 
AD -BD = AC' -BC' 
AD -AC' = BD-BC'

Now, if D is between C' and B (or D is at or beyond B) then AD > AC' and BD < BC', which would give the left and right sides of the last equation different signs.  Likewise, if D is between C' and A (or D is at or beyond A) then AD < AC' and BD > BC', which is similarly impossible, so D=C'.

Note: There is nothing about either direction of this proof that precludes points A', B', and/or C' from being beyond the vertices of the triangle.  That is, point A', for example, may be on the extension of side BC, rather than on the BC itself.  In fact, as cut-the-knot points out, points A', B', and C' need not be even on the sides of the triangle, but rather anywhere on a line perpendicular to the corresponding side of a triangle.  Even in this case, Carnot's Theorem still holds.

Corollary 1: Perpendicular bisectors of the sides of a triangle concur in a point (the circumcenter.)

Since the perpendiculars bisect AB, BC, and CA,

AC' = BC', BA' = CA', and CB' = AB',
so AC'² = BC'² , BA'² = CA'² , and  CB'² - AB'² , yielding immediately the result that
AC'² - BC'² + BA'² - CA'² + CB'² - AB'² = 0, so by Carnot's theorem, the three perpendiculars concur on a point.

Corollary 2: The altitudes of a triangle concur in a point.

Letting AA', BB', and CC' be the altitudes of ABC, we see that each altitude is a leg of two right triangles, so that

CC'² + AC'² = AC²,
-CC'² - BC'² = -BC²,
AA'² + BA'² = AB²,
-AA'² - CA'² = -AC²,
BB'² + CB'² = BC²,
-BB'² - AB'² = -AB²

Now, just as we did to prove Carnot's theorem, we simply add up the six equations to get

AC'² - BC'² + BA'² - CA'² + CB'² - AB'² = 0, so by Carnot's theorem, the three altitudes concur on a point.

. . . . . . .  A page on Menelaus' Theorem would be nice, too. 

. . . . . . A page on Stewart's Theorem (  ) would be nice, too

Posted on Saturday, 25 August, 2007 - 02:35 am:   
"P, Q, R are points on the sides BC, CA, AB of triangle ABC. Prove that the perpendiculars to the sides at these points meet at a common point iff:

BP� + CQ� + AR� = PC� + QA� + RB�."

I've tried using pythagoras a fair few times and have combined some equations, but I'm not really getting anywhere.

Any help ?

Thanks a lot.

Posted on Saturday, 25 August, 2007 - 05:45 am:   
I wonder if it's related somehow to Menelaus' Theorem. Honestly, I don't know, but when I draw the picture for your question, I am struck by the fact that the three perpendiculars are transversals of the triangle, and so Menelaus applies to each one. Then, perhaps, some identities can be derived, and things will start to cancel.

For more about Menelaus, see Dr. Math. Also, see the really beautiful Geometry from the Land of the Incas.

Edit: Aha! Carnot's Theorem. (But I still urge you to explore the Land of the Incas!)


Internet references is a really easy-to-follow proof, from which I borrowed liberally here. has a java-enabled diagram on which you can drag the vertices of the triangle, and watch the various similar triangles in action.

Related pages in this website:

Summary of geometrical theorems

Triangles -- a home page for anything about triangles.

Circumcenter, one of the triangle centers.

Isogonal Conjugates -- the relationship between the orthocenter and circumcenter of a triangle

The webmaster and author of this Math Help site is Graeme McRae.