Two right triangles, △ABD and △ACD, were created from △ABC by constructing AD ⊥ CB.
Follow these steps to prove that #&\frac {\sf sin(B)}{b} = \frac {\sf sin(C)}{c}#&.

✓ 1. For △ABD, sin(B) = c/h h/c

✓ 2. Multiply both sides of the equation by c and simplify. The equation sin(B) = #&\frac {h}{c}#& becomes c sin(B) = h h sin(B) = h .

✓ 3. For △ACD, sin(C) = b/h h/b

✓ 4. Multiply both sides of the equation by b and simplify. The equation sin(C) = #&\frac {h}{b}#& becomes b sin(C) = h h sin(C) = h .

✓ 5. By the multiplication substitution property, we can set the expressions equal to each other.
                          c sin(B) = b sin(C)

✓ 6. Divide both sides of the equation by b c bc and simplify to get #&\frac {\sf sin(B)}{b} = \frac {\sf sin(C)}{c}#&.