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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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