Law of Sines

Theorem
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

Where $$a$$, $$b$$, and $$c$$ are the sides opposite $$A$$, $$B$$ and $$C$$ respectively.

Proof 1
Construct the altitude from $$B$$.



It can be seen from the definition of sine that $$\sin A=\frac{h}{c}$$ and $$\sin C=\frac{h}{a}$$.

Thus $$h=c\sin A$$ and $$h=a\sin C$$.

This gives us $$c\sin A=a\sin C$$.

So $$\frac{a}{\sin A}=\frac{c}{\sin C}$$.

Similarly, constructing the altitude from $$A$$ gives us $$\frac{b}{\sin B}=\frac{c}{\sin C}$$.

Proof 2
Construct the circumcircle of $$\triangle ABC$$, let $$O$$ be the circumcenter and $$R$$ be the circumradius.

Construct $$\triangle AOB$$ and let $$E$$ be the foot of the altitude of $$\triangle AOB$$ from $$O$$.



Now we have

$$\angle ACB = \frac{\angle AOB}{2}$$ by the inscribed angle theorem

$$AO = BO$$ from the definition of the circumcenter and $$\angle AEO = \angle BEO$$ from the definition of altitude and the fact that all right angles are congruent

Therefore $$AE = BE$$ from the Pythagorean Theorem, and then $$\angle AOE = \angle BOE$$ from Triangle Side-Side-Side Equality.

This gives us $$\angle AOE = \frac{\angle AOB}{2}$$

So $$\angle ACB = \angle AOE$$

Then $$\sin C = \sin (\angle AOE) = \frac{\frac{c}{2}}{R}$$ by the definition of sine

$$\frac{c}{\sin C}=2R$$

Because the same statement holds for all three angles in the triangle, $$\frac{c}{\sin C}=2R=\frac{b}{\sin B}=2R=\frac{a}{\sin A}$$.

Note that this proof also yields a useful extension of the law of sines: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}=2R$$