Pythagoras's theorem proof (rational trigonometry)

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The Pythagorean theorem, expressed as a relation between the quadrances of the sides of a right triangle, is one of the five basic laws of the rational trigonometry system devised in the early 2000s by Dr. Norman Wildberger.

There are many classical proofs of Pythagoras's theorem; this one is framed in the terms of rational trigonometry.

The spread of an angle is the square of its sine. Given the triangle ABC with a spread of 1 between sides AB and AC,

$Q(AB) + Q(AC) = Q(BC)\,$

where Q is the "quadrance", i.e. the square of the distance.

[edit] Proof

Illustration of nomenclature used in the proof.

Construct a line AD dividing the spread of 1, with the point D on line BC, and making a spread of 1 with DB and DC. The triangles ABC, DBA and DAC are similar (have the same spreads but not the same quadrances).

This leads to two equations in ratios, based on the spreads of the sides of the triangle:

$s_C = \frac{Q(AB)}{Q(BC)} = \frac{Q(BD)}{Q(AB)} = \frac{Q(AD)}{Q(AC)}.$

$s_B = \frac{Q(AC)}{Q(BC)} = \frac{Q(DC)}{Q(AC)} = \frac{Q(AD)}{Q(AB)}.$

Now in general, the two spreads resulting from dividing a spread into two parts, as line AD does for spread CAB, do not add up to the original spread since spread is a non-linear function. So we first prove that dividing a spread of 1, results in two spreads that do add up to the original spread of 1.

For convenience, but with no loss of generality, we orient the lines intersecting with a spread of 1 to the coordinate axes, and label the dividing line with coordinates $(x 1, y 1)$ and $(x 2, y 2)$ . Then the two spreads are given by:

$s_1 = \frac{(x_2 - x_2)^2 + (y_2 - y_1)^2}{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \frac{(y_2 - y_1)^2}{(x_2 - x_1)^2 + (y_2 - y_1)^2},$