^{2}+ 4

^{2}=5

^{2}set next to 3

^{3}+4

^{3}+5

^{3}=6

^{3}and wondered.... Could it be????... Is it possible???? Oh go on, you know you are going to check... how could you resist?

Just reminded of some other "almost Pythagorean" relations in geometry from an old (1932) Mathematics Teacher article...(Thank you, again, Dave Renfro)

The Two Triangles formed when the median is drawn to any side are

**almost Pythagorean**

Treat the two sides not cut by the median as if they were the hypotenii (hypotenuses?) of a right triangle. Both are wrong, but in sum, they are right..

That is, while neither of the following are true, AM

^{2}+MC

^{2}= AC

^{2}

AM

^{2}+MB

^{2}=AB

^{2}

Adding the two equations produces a true relation.

AM

^{2}+MC

^{2}+AM

^{2}+MB

^{2}= AC

^{2}+AB

^{2}

The proof is pretty easy using the Law of Cosines.

A second, and about as easy to prove.... Let G be the centroid of Triangle ABC, then AB

^{2}+BC

^{2}+CA

^{2}= 3 (GA

^{2}+GB

^{2}+GC

^{2})

A third, which is, I believe, both necessary and sufficient to prove a parallelogram is that the sum of the squares of the diagonals is equal to the sum of the squares of the sides.

And if it is not a parallelogram, but is a trapezoid, then the sum of the squares of the diagonals is equal to the sum of the squares of the non-parallel sides, plus twice the product of the parallel bases (sort of a law-of-cosines look-a-like).

Recently saw the old Wizard of Oz mis-spoken statement of the Theorem by Ray Bolger as the Scarecrow posted at 360, so I thought this was a good time to include it for my students (and anyone else who has never seen it, or just wants to chuckle one more time). [] Play it for your students and see if they can catch all the mistakes...