Carnot's Theorem

Carnot's theorem is a direct application of the theorem of Pythagoras. It's a nice little theorem that is unlikely to be rediscovered with dynamic geometry software. A penchant for doodling and imagination this is all that is probably needed to rediscover a theorem like that.

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

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

Proof

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-angled triangles, three pairs of which share a hypotenuse, whilst three other pairs share a leg with a vertex at P. The situation begs for an application of 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.

Taking the first, third, and fifth equations with the sign "+", and the second, fourth, and sixth equations with the sign "-", and adding up all six leads, after obvious simplifications, to (1).

To proof the converse, assume (1) and let P be the point of intersection of, say, A'P and B'P. Drop a perpendicular DP from P to AB. By the already proven part:

AD² - BD² + BA'² - CA'² + CB'² - AB² = 0,

so that

(2)	AD² - BD² = AC'² - BC'².

This is only possible if D = C'. Why so. Let BD = x, AB = c. Then AD = c - x, and

AD² - BD² = (c - x)² - x² = c² - 2cx,

which is a linear function of x: f(x) = c² - 2cx. f therefore is 1-1. In other words, f(x) can't have the same value for two different x's. This means that (2) does imply D = C'.

We are naturally coming to a related problem: Given two points A and B. What is the locus of points M such that AM² - BM² is constant?

Let M be a point on the locus and H the foot of the perpendicular from M to AB. By the Pythagorean theorem:

AM² - AH² = MH² = BM² - BH²,

i.e.,

AM² - BM² = AH² - BH²,

which means that, if M lies on the locus in question, so does H. And vice versa. Since the same H serves as the foot of the perpendicular MH for all points M of the line perpendicular to AB at H, this line solves the problem.

If we place the origin of a Cartesian coordinates system at B with the x-axis along AB and the y-axis perpendicular to it, we may assign coordinates (x, y) to point M. Then

BM² = x² + y², whereas
AM² = (c - x)² + y².

Thus, as above,

AM² - BM² = (c - x)² + y² - x² - y² = f(x),

where f(x) is the function defined above. If x is fixed, so is f(x), and also AM² - BM², as expected. From this fact, or from the underlying proof, it is clear that, for (1) to hold, the points A', B' C' need not be necessarily located on the sides of ΔABC.

Corollary

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

Indeed, AC' = BC', BA' = CA', and CB' = AB' imply (1).
The altitudes of a triangle concur in a point.

Indeed, if AA', BB', and CC' are altitudes of ΔABC, then

AB² - BA'² = AA'² = AC² - CA'²,

so that

AB² - AC² = BA'² - CA'²,

Similarly,

AC² - BC² = AC'² - BC'² and
BC² - AB² = CB'² - AB'², which,

when added, give (1).
The perpendiculars to the sides of ΔABC at the points where the incircle meets the sides are concurrent (at the incenter.)

Obvious.
The perpendiculars to the sides of ΔABC at the points where the excircles meet the sides are concurrent.

Indeed [F. G.-M., p. 555], in this case we have

CA' = AC', BA' = AB', and BC' = CB',

which implies (1).
Existence of Orthopole

To every line m in the plane of ΔABC there correspond a point - the orthopole of M with respect to ΔABC. To define the point, first drop perpendiculars AL, BM and CN from the vertices of ΔABC onto m. From the three points thus obtained drop perpendiculars on the "opposite" sides of the triangle: from L onto BC, from M onto AC, and from N onto AB. The latter three lines are concurrent, and the point where they meet is known as the orthopole of m with repsect to ΔABC.

Darij Grinberg came up with an ingenious proof of this result based on Carnot's theorem (1). As we already know, the points A', B', C' need not be located on the sides of ΔABC. So let's take A' = L, B' = M, and C' = N. Applying the Pythagorean theorem several times, we obtain

AM² - LM² = AL² = AN² - LN²,

so that

AM² - AN² = LM² - LN².

Similarly,

BN² - BL² = MN² - LM², and
CL² - CM² = LN² - MN²,

which, when added, produce

AM² - AN² + BN² - BL² + CL² - CM = 0.

References

F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991

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