Law of Sines/Proof 3

Acute Case
Let $\triangle ABC$ be acute.


 * Law-of-sines-proof-3.png

Construct the circumcircle of $\triangle ABC$.

Let its radius be $R$.

Construct its diameter $BX$ through $B$.

By Thales' Theorem, $\angle BAX$ is a right angle.

From Angles in Same Segment of Circle are Equal:
 * $\angle AXB = \angle ACB$

Then:

The same construction can be applied to each of the remaining vertices of $\triangle ABC$.

Hence the result.

Let $\triangle ABC$ be obtuse.


 * Law-of-sines-proof-3-obtuse.png

As for the acute case, construct the circumcircle of $\triangle ABC$.

Let its radius be $R$.

Construct its diameter $BX$ through $B$.

By Thales' Theorem, $\angle BAX$ is a right angle.

We note that $\Box ABXC$ is a cyclic quadrilateral.

From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles:
 * $\angle BAC = 108 \degrees - A$

Hence using a similar argument to the acute case:

and the result follows.