While testing out the geometer's sketchpad, I came across a Tangram proof of the theorem of Pythagoras. I assume that it must be one of umpteen thousand proofs. However, not being a geometer, I have no idea if it is in fact wellknown.
Let me draw the picture:
B' C A' _______ __________ | | C, C' are right angle vertices | | | | A | | | | B | | D' |_____ ___________| D
ACB is the given right triangle with C the right angle vertex. AB'C and CA'B as well as AC'B are 45-45 right triangles by construction. AD'C' and C'DB are right triangles which can be checked to be congruent and similar to ACB with ratio 1:sqr(2)
B'A'DD' is therefore a square with sides of lengths:
(a + b)/sqr(2)
I need the special case of the Pythagoras theorem that the hypoteneuse of a 45-45 right triangle is sqr(2) times the length of one leg. This can be given a Tangram proof. Namely two of these form a square with the hypoteneuse glued to each other. However, two of these may be glued along a common leg to form a 45-45 right triangle so that 4 of them form a square whose side is equal to the original hypoteneuse. An area calculation for squares gives us the desired special case.
For right triangles, the area is half the product of the two perpendicular legs. Now, the displayed square has area:
(a + b)^2 * (1/2)
It is the sum of 6 right triangles. Three are 45-45 right triangles with hypoteneuses a, b, and c by contruction. Of the remaining 3, one is the original one with area ab/2 while the other two are congruent to each other and simiar to the given one with ratio sqr(2):1, thus have areas equal to ab/4. Adding up, we get a total area of:
(a^2 + b^2 + c^2)/2 + ab/2 + ab/4 + ab/4
Equating the two different calculations, multiplying by 2 on both sides and simplify, we get the theorem of Pythagoras:
a^2 + b^2 = c^2
Note: Coxeter (Intro. to Geometry) gives a proof that involves dropping the perpendicular from C to AB and uses proportionality principle for similar triangles and their areas. It is compact and neat, but the idea of a square is hidden. The Tangram proof does not involve squares, but it involves a number of half squares (from a half-wit (!)).