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Is there not a proof using similar triangles somehow? Because the whole 'sine' part seems like a bit of a red herring, they're basically just considering a coup
by contravariant 4y ago
Is there not a proof using similar triangles somehow? Because the whole 'sine' part seems like a bit of a red herring, they're basically just considering a couple of ratios between different lengths, they do not use any properties of the sine function as such (in particular it does not look like they're using sin(x)^2 + cos(x)^2 = 1, which would make the proof trivial)
- chr1 4y agoYes, if they droped height h to the side of isoscales triangle, then from similar triangles they would have h/A=c/C and h/b=2a/c and would get the result without mention of sine rule or areas.