Extending from the proportional center “P” of right triangle ABC, the application of bn=(.5ab+2(a2+b2))b.5a{b}_n=\sqrt{\frac{\left(.5ab+2\left({a}^2+ {b}^2\right)\right) {b}}{.5a}} and an=(ab)bn{a}_n=\left(\frac{ {a}}{b}\right) {b}_n yield quadrilaterals, being trapezia (trapezoids, as your convention dictates) of area corresponding to the square of their respective sides… ad infinitum. The area of BCC`B` = The square of side a The area of ACC`A` = The square of side b The area of ABB`A` = The square of side c While dissimilar, the trapezia preserve both the triangle proportionally and Pythagorean (et al.) Theorem conformity.
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