Pythagorean Theorem via an Isosceles Triangle

The proof of the Pythagorean Theorem illustrated by the applet below is due to Gaetano Speranza:

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

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 a + m and m, where m is half BB' - BC = c - a:

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).

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