Steve's new proof of the Pythagorean Theorem! Feb 3rd,
2014. My homepage!
Feb 21: Oh no! I just discovered that Albert Einstein
this exact proof when he was 12. Since I am 5 times that
age, my IQ must be 1/5th his.
Given: a right triangle, sides of length a and b, hypotenuse length
c. Prove: c2 = a2 + b2.
Draw a line perpendicular to the hypotenuse and thru the opposite
corner point. This line divides the triangle into two smaller
triangles. To wit:
Note that the two smaller trianges are both similar to each other
and also similar to the original triangle. (Each shares a common angle and a
Note that the area of the original triangle equals the sum of the
areas of the two new triangles (obviously, by construction).
Now, let's draw some kind of a "bump" anywhere on the original
triangle, of any arbitrary shape, and then draw a similar bump on
both of the other triangles ("similar" in the rigorous sense,
identical shape, proportionally sized at the same proportion as the
Claim: The area of any bump on the original triangle equals the sum
of the areas of the similar bumps on the other two! That is:
(This should also be "obvious": The sum of the areas of the two
triangles equals the area of the larger one, by construction. And
since they are all scaled by the same ratios, if it holds true for
the triangles, it must also hold true for the bumps!)
Now, if you believe that simple truth, then the final step is quite
trivial, and you have what I think is the world's most trivial,
simple, and easy to understand proof ever of the Pythagorean
Drum roll please.... I'm now gonna draw a very specific "bump" on
all three triangles. Namely, a square on their hypotenuses:
Do you get it already? QED!!
These squares are just big "bumps", no?
And we just said:
"The area of any bump on the original triangle equals
the sum of the areas of the similar bumps on the other two!".
Thus, since a "square" is a "bump", and the above statement holds
for any bump, then:
The area of the square on the hypotenuse of the original
triangle equals the sum of the areas of the (similar) squares on
the hypotenuses of the other two
QEF'nD: Quod erat f'n demonstrandum!
I'm gonna claim the above is the simplest proof of the Pythagorean
Theorem ever devised! It is almost completely obvious by inspection.
If you delete all text on this page and leave
just the drawings, the sequence of figures almost proves
itself! No algebra. No determination of the length of the
perpendicular, or the relative lengths of the two pieces c is broken
into. No nothin'!
In thinking more on this, I have come to this new idea, which I have
not seen anywhere else. And it is this: is there any other 2D shape
which can be cut with a single cut and produce two pieces which are
similar to each other AND ALSO similar to the original? For example,
cutting a rectangle in half produces two similar halves, but they
are not similar to the original rectangle. Or cutting a square off
of a golden rectangle produces one piece similar to the original but
not to the other piece.
Thus, Swift's Right Triangle Hypothesis:
A right triangle is the ONLY 2D shape which can be cut
with a single straight-line cut such that the two new pieces are
both similar to each other and also similar to the original shape.
And THIS is what is unique about right triangles. This is the more
fundamental truth from which the Pythagorean Theorem is merely a