I need an algebric proof of the theorem of similar triangles (C/c=A/a=B/b).
For two similar triangles, that A/a=C/c=B/b.TD said:Proof of what exactly?
Well trigonometric functions don't proove the theorem, they are a product of it so it is not really a proof... Anyway I'm putting my mind into it and I hope to find why it is so.mathmike said:wouldnt the proof be something like this
Sin A / a = Sin B /b = Sin C / c