You should draw a triangle ABC inscribed in a circle whose radius is r. You need to draw a diameter that starts from the vertex A and it ends in the point D.
Joining the points A, B, D, you'll get a right triangle since hat B is opposed to diameter AD.
You need to express the sine of the angle `hat(ADB) ` such that:
`sin hat(ADB) = (AB)/(AD)`
Notice that `AD` represents the diameter of circle, hence `AD = d = 2r.`
Notice that the inscribed angles `hat(ACB) ` and `hat(ADB)` cut the same chord, hence `hat(ACB) = hat(ADB).`
Since `sin hat(ADB) = (AB)/(AD) => sinhat(ACB) = (AB)/(AD) = (AB)/(d) => sin hatC = (AB)/(d) `
You may come up with the following substitutions for the sides of triangle ABC such that:
`AB = c` (opposed to `hatC` )
`AC = b` (opposed to `hatB` )
`BC = a (opposed hatA)`
Hence, usiing these new notations yields:
`sin hatC = c/(d) => c/(sin hat C) = d`
Resoning by analogy yields:
`c/(sin hat C) = a/(sin hat A) = b/(sin hat B) =d`
Hence, using the sine formula and circle identities you may prove the law of sines such that: `c/(sin hat C) = a/(sin hat A) = b/(sin hat B) = d.`