8.3 SURVIVAL GEOMETRY II 273
The beautiful crux idea is an auxiliary line m drawn through the right angle (B), per
pendicular to the hypotenuse AC. Line m divides the big square-on-the-hypotenuse
into two rectangles, RSIA and RCHS. If we slide the line segment RS along m, these
rectangles will shear into parallelograms, but their areas will not change.
In particular, we can slide RS so that R coincides with B. Meanwhile, we can play
a similar game with the two smaller squares-on-the-legs. If we shear BAFG into a
parallelogram by sliding segment FG straight down to m (so that G lies on m), the area
will not change. Likewise, shear BDEC by sliding DE to the left. The result is below.
Now it's obvious-the two "leg" squares, and the two "hypotenuse" rectangles sheared
into congruent parallelograms! So of course the areas are equal! _
8.3.6 OK, it's not quite obvious. Prove rigorously that the parallelograms are indeed
congruent. You'll need to do a little angle chasing to get congruent copies of the
original triangle.
The Pythagorean Theorem has literally hundreds of different proofs. Perhaps the sim
plest, discovered in ancient India, uses nothing but "cutting and pasting."
Example 8.3.7 A "dissection" proof of the Pythagorean Theorem. The picture below
should make it clear. The area of the first square is a^2 + b^2 plus four times the area of
the right triangle. The area of the second square is c^2 plus four times the area of the
right triangle. Since the squares are congruent, their areas are equal, so we conclude
that a^2 + b^2 = c^2. _
