1. ∠BOD = ∠α
2 . sin α = opp/hyp = (a/2)/R = a/[2R] or, stated another way , 2R = a/sinα (1)
3. Bisect AC at E and draw OE perpendicular to AC...thus ∠OEC will be a right angle...and, by SSS, triangle AOE = triangle COE.....thus angle AOE = angle COE .. ...let ∠ABC = ∠β and let AC = b......which implies that EC = b/2
Then, 1/2∠COA = ∠ABC.....but, since ∠COE = 1/2∠COA ....then ∠COE = ∠ABC = ∠β
And, as before....... sin COE = sin β = (b/2)/OC = (b/2)/R = b/[ 2R] which implies that
2R = b/sinβ (2)
Finally.......bisect AB at F and draw OF perpendicular to AB.......then ∠OFA will be a right angle....and,by SSS, triangle AOF= BOF......thus ∠AOF =∠ BOF......let ∠ACB = ∠c and let AB = c........which implies that FA = c/2
Then, 1/2∠AOB = ∠ACB....but, since ∠AOF = 1/2∠AOB, then ∠AOF = ∠ACB = ∠c
As before........sin AOF = sin c = (c/2)/OA = (c/2)/R = c/ [ 2R] which implies that
2R = c/sin c (3)
So
2R = 2R = 2R means that
(1) = (2) = (3) and we have that
a/sinα = b/sinβ = c/sin c [ the Law of Sines]
