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 https://www.java.com/en/download/index.jsp, 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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