Referring to TriangleIdentity by 伍柒貳 a while ago, considering $\bigtriangleup$ ABC, it is proved that:
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C (1*) $$
I want to take angle $A = 0$ to look at the angle relationship of parallel lines so created.
From the above $ \cos B + \cos C = 0 $ gives two solutions as :
$$ 2 \cos( \frac {B+C}{2} ) \cos( \frac{B-C}{2} ) = 0 \, (2*) $$
$$ B + C = \pi ; B - C = \pi; (3*)$$
With $ A + B + C = \pi $ the first gives Euclid's proposition ( Two parallel lines cut by any transversal, two angle sum on one side equals $ \pi$) as expected.
But the second part of (3*) gives
$$ B = \pi - A/2 , \, C = -A/2 $$
The second situation has me stumped. Is the Sine & Cos Triangle Law someway flawed? Is there a seed in it of hyperbolic geometry? Please comment on any related aspect. Regards
There are three "triangle"s with $\angle A = 0 $ and $AB $ parallel $AC$
a "triangle" with $\angle A = 0 $, $\angle B = 0 $ , $\angle C = \pi $
a "triangle" with $\angle A = 0 $, $\angle B = \frac{\pi}{2} $ , $\angle C = \frac{\pi}{2} $
a "triangle" with $\angle A = 0 $, $\angle B = \pi $ , $\angle C = 0 $