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Induction isn't so bad. I'm not sure how they're getting away with the Law of Sines, though. The usual proof of LoS that I know is dependent on the existence of

by scythe 4y ago

Induction isn't so bad. I'm not sure how they're getting away with the Law of Sines, though. The usual proof of LoS that I know is dependent on the existence of the circumcircle. But I don't know how to prove the existence of the circumcircle without dragging in a lot of geometry. Or you can use the area formula, which makes the proof similar to other arguments that use the area formula.

SamReidHughes 4y ago
There is a very straightforward proof that involves dropping an altitude. https://www.mathopenref.com/lawofsinesproof.html https://www.mathopenref.com/lawofsinesproof.html You could use a similar technique to make this proof not reference trigonometric functions.
- crdrost 4y ago
  For that matter the easiest proof of Pythag that I know of involves dropping an altitude: Look at the right triangle the normal way up, clearly the area of the triangle is k c² (k = ½ sin α sin β if you like, but it just matters that it's the same nonzero k for all similar triangles). Now roll it onto its hypotenuse, drop an altitude, and observe that both subtriangles are similar to the first one, kc² = ka² + kb². The diagram that's a bit involved is the angle sum diagram, you start with a right triangle (a,b,c) with some angle α, extend it to a new triangle (a,b', c') with angle α+β, then make the new triangle with angle β that you stacked on top of the original triangle into a right triangle with angle β (c', d, e) by extending the hypotenuse of the (a,b,c) triangle to a point P, basically until the angle with the hypotenuse c' is 90°. Drop a dotted line to the x-axis from P and you can work out that the dotted line is at x=cos α cos β, and its distance to a is sin α sin β. Similarly the y-coordinates give sin α cos β + cos α sin β. As you say, you can do all of this without angles except for defining the first triangle with angle α+β, which you might not even need... We just need it here for sin(2 α) which is something like reflecting the same triangle about its hypotenuse?
  - alex_smart 4y ago
    For this proof to be a proof, you first have to define what area is and why it should scale as kc² and why the sum of areas of the two the smaller triangles should equal the area of the larger triangle.
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 ago
  Yes, 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.

by scythe 4y ago

SamReidHughes 4y ago

There is a very straightforward proof that involves dropping an altitude. https://www.mathopenref.com/lawofsinesproof.html https://www.mathopenref.com/lawofsinesproof.html You could use a similar technique to make this proof not reference trigonometric functions.

crdrost 4y ago
For that matter the easiest proof of Pythag that I know of involves dropping an altitude: Look at the right triangle the normal way up, clearly the area of the triangle is k c² (k = ½ sin α sin β if you like, but it just matters that it's the same nonzero k for all similar triangles). Now roll it onto its hypotenuse, drop an altitude, and observe that both subtriangles are similar to the first one, kc² = ka² + kb². The diagram that's a bit involved is the angle sum diagram, you start with a right triangle (a,b,c) with some angle α, extend it to a new triangle (a,b', c') with angle α+β, then make the new triangle with angle β that you stacked on top of the original triangle into a right triangle with angle β (c', d, e) by extending the hypotenuse of the (a,b,c) triangle to a point P, basically until the angle with the hypotenuse c' is 90°. Drop a dotted line to the x-axis from P and you can work out that the dotted line is at x=cos α cos β, and its distance to a is sin α sin β. Similarly the y-coordinates give sin α cos β + cos α sin β. As you say, you can do all of this without angles except for defining the first triangle with angle α+β, which you might not even need... We just need it here for sin(2 α) which is something like reflecting the same triangle about its hypotenuse?
- alex_smart 4y ago
  For this proof to be a proof, you first have to define what area is and why it should scale as kc² and why the sum of areas of the two the smaller triangles should equal the area of the larger triangle.

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 ago
Yes, 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.