Consider Triangle ABC is inscribed in a circle with Centre O.

From above figure,

Radius = OA = OB = OC = r

Let the sides of the triangle are

a = BC , b = AC , c = AB

Angles are A , B , C.

From OBD right angled triangle

sinA = BD/OB

sinA = (a/2)/r

sinA = a/2r

(sinA)/ a = 1/2r         ------------------------(1)

From OCE right angled triangle

sinB = CE/OC

sinB = (b/2)/r

sinB = b/2r

(sinB)/ b = 1/2r         ------------------------(2)

From OAF right angled triangle

sinA = AF/OA

sinC = (c/2)/r

sinC = c/2r

(sinC)/ c = 1/2r         ------------------------(3)

From (1) , (2) , (3) Equations

(sinA)/ a = (sinB)/ b = (sinC)/ c = 1/2r