
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.
Proof
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' = BDBC'
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 cuttheknot 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 ( http://mathforum.org/library/drmath/view/54814.html ) would be nice, too
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. ~~Simba 

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!) 
http://www.cuttheknot.org/pythagoras/Carnot.shtml is a really easytofollow proof, from which I borrowed liberally here.
http://www.cuttheknot.com/Generalization/ceva.html has a javaenabled diagram on which you can drag the vertices of the triangle, and watch the various similar triangles in action.
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.