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Consider any triangle ABC, and let AD be the altitude from A on to BC.

Then sin(B) = AD/AB so that AD = AB*sin(B)

and sin(C) = AD/AC so that AD = AC*sin(C)

Therefore AB*sin(B) = AC*sin(C)

or c*sin(B) = b*sin(C) where the lower case letter represents the side opposite the angle with the upper case name.

Divide both sides by bc to give sin(B)/b = sin(C)/c.

Similarly, using the altitude from B you can show that sin(A)/a = sin(C)/c.

Combining with the previous result,

sin(A)/a = sin(B)/b = sin(C)/c.

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