# 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 Q_{a}, Q_{b}, Q_{c} on the sides of ΔABC. (I'll use the same symbols to denote their areas.)

By the construction, h = b/2, so that Q_{b} = 4h².

Q_{c} is also the area of the square with side BB'. It follows that

Q_{c} | = Q_{a} + 4R + 4Q | |

= Q_{a} + 4am + 4m² | ||

= Q_{a} + 4h² | ||

= Q_{a} + Q_{b}. |

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).

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

70768184