3 ms·

There is a very straightforward proof that involves dropping an altitude. https://www.mathopenref.com/lawofsinesproof.html https://www.mathopenref.com/lawofsin

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.

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.