Spread Law

The Spread Law states that for any triangle with non-zero quadrances

$$ \frac{s_1}{Q_1} = \frac{s_2}{Q_2} = \frac{s_3}{Q_3} $$

Proof
To simplify the next calculation let

$$a_1 \equiv (y_1 - y_2)$$ $$b_1 \equiv (x_2 - x_1)$$ $$a_2 \equiv (y_1 - y_3)$$ $$b_2 \equiv (x_3 - x_1)$$ $$a_3 \equiv (y_2 - y_3)$$ $$b_3 \equiv (x_3 - x_2)$$

Suppose that three points $$A_1 \equiv [x_1, y_1]$$, $$A_2 \equiv [x_2, y_2]$$, and $$A_3 \equiv [x_3, y_3]$$ form a right triangle. Then

$$ A_1 A_2 = \langle a_1 : b_1 : x_1 y_2 - x_2 y_1 \rangle $$

$$ A_1 A_3 = \langle a_2 : b_2 : x_1 y_3 - x_3 y_1 \rangle $$

$$ A_2 A_3 = \langle a_3 : b_3 : x_2 y_3 - x_3 y_2 \rangle $$

Starting from the definition of spread

$$ s(A_1 A_2, A_1 A_3) = \frac{(a_1 b_2 - a_2 b_1)^2}{(a_1^2 + b_1^2)(a_2^2 + b_2^2)}$$ $$= \frac{(a_2 b_3 - a_3 b_2)^2 + (a_2 a_3 + b_2 b_3)^2}{(a_1^2 + b_1^2)(a_2^2 + b_2^2)}$$

the last equality holds as $$ a_2 a_3 + b_2 b_3 = 0 $$ becaues $$ A_1 A_3 $$ and $$ A_2 A_3 $$ are perpendicular. Now

$$\frac{(a_2 b_3 - a_3 b_2)^2 + (a_2 a_3 + b_2 b_3)^2}{(a_1^2 + b_1^2)(a_2^2 + b_2^2)}$$ $$= \frac{(a_2 ^ 2 + b_2 ^ 2) (a_3 ^ 2 + b_3 ^2)}{(a_1^2 + b_1^2)(a_2^2 + b_2^2)}$$ $$= \frac{Q_2 Q_1}{Q_3 Q_2} = \frac{Q_1}{Q_3}$$

Now that spread is related to the ratio of quadrants the strategy from the traditional proof for the law of sine suffices to prove the spread law.