Pythagorean Theorem via an Isosceles Triangle
The proof of the Pythagorean Theorem illustrated by the applet below is due to Gaetano Speranza:
|What if applet does not run?|
Given right ΔABC, extend side BC to B' so that BB' = AB, the hypotenuse of the triangle. ΔABB' is isosceles, so that the altitude BH to AB' is also the median: H is the midpoint of AB'.
ΔBHB' is right at H so that the square of its altitude from H, which I'll denote h, equals the product of the segments of the hypotenuse which are equal to
h² = (a + m)m = am + m².
Form squares Qa, Qb, Qc on the sides of ΔABC. (I'll use the same symbols to denote their areas.)
By the construction, h = b/2, so that Qb = 4h².
Qc is also the area of the square with side BB'. It follows that
|Qc||= Qa + 4R + 4Q||= Qa + 4am + 4m²||= Qa + 4h²||= Qa + Qb.|
Gaetano observes that the only areas that are evaluated in the proof are those whose sides are parallel (and perpendicular) to the legs of ΔABC, making it rather "analytically" (="cartesian-wise") tasteful. This is achieved by mapping the only slanted element of the triangle - the hypotenuse AB - on one of the legs (read one of the axes).
Copyright © 1996-2018 Alexander Bogomolny