Proving the Sine Rule with one line. Working on a general proof of the Law of Sines for ALL Euclidean triangles. Right triangles are easy. Acute triangles are just two proofs of the right triangle. But this is not sufficient for me. I want a one time proof for all Euclidean triangles. That brings me to the obtuse case. The first step is to draw the perpendicular to the base. This gives me two right triangles so I can prove the ratios for the opposing angles and sides. So that:
$$\frac {\sin\alpha} {a}=\frac{\sin\gamma}{c} $$
(that is the easy part)
My problem is proving that the sine of the two halves of angle $\beta$ call them $\beta_1$ and $\beta_2$ over the two halves of the bisected side (call them $b_1$ and $b_2$) satisfy the proof. So that : $$\frac {\sin \alpha}{a} = \frac{\sin\gamma}{c}= \frac {\sin(\beta_1 + \beta_2)}{b_1+b_2}$$ with out resorting to drawing more lines or using the ambiguous case. 

Draw a Circum-circle around the triangle and a diameter that passes through intersection of $a$  and $c$. Draw a cyclic quadrilateral including the endpoint of the diameter by joining lines of the triangle opposite to the given triangle.Mark angles $\alpha $ and $ \gamma$ in the opposite segment as shown.It is clear that for all positions of right side vertex $ \gamma+ \alpha $ which is constant on the arc
$$ \frac {\sin\alpha} {a}=\frac{\sin\gamma}{c} = \frac{\sin(\beta_1+\beta_2)}{(b_1+b_2)}= \frac{1} {CircumDiameter D} $$
is constant, the reciprocal diameter D.

