Dissimilar Pythagorean Shapes via the Vanishing Point of a Right Triangle | Steve Wait – May 2, 2022

Extending from the proportional center “P” of right triangle ABC, the application of

${b}_n=\sqrt{\frac{\left(.5ab+2\left({a}^2+ {b}^2\right)\right) {b}}{.5a}}$

and

${a}_n=\left(\frac{ {a}}{b}\right) {b}_n$

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.