In the acute angle △AB=c, let BC=a, AC=b and AB=c. For CH⊥AB, the foot is the point H?

CH=a sinB？ CH=b Sina? ∴a？ Get? a/sinA=b/sinB？ Similarly, in △ABC,? b/sinB=c/sinC？ Step two. ? Prove that a/sinA=b/sinB=c/sinC=2R:? As shown in the figure, any triangle ABC is the circumscribed circle of ABC. Make the diameters BD and D cross ≧O and connect DA. ? Because the circumferential angle of the inner diameter of the same circle or the same circle is a right angle ∠DAB=90 degrees? Because the circumferential angles of the same arc in the same circle or equal circle are equal, ∠D equals ∠ C? So c/sinC=c/sinD=BD=2R.