Easy foorp

That proof can also be done sort of backwards and is arguably easier that way. Instead of using three triangles of known dimensions and proving they can be assembled into a rectangle, start with the rectangle and find the dimensions of the triangles.

Take a right triangle with sides ab, and c. Draw perpendiculars to the hypotenuse at each end and a parallel line through the third vertex to make a rectangle BCDE.

Angles CDA and EAD are congruent (alternate interior angles). So are ADE and ACD since both are complementary with ADC. So triangles ADE and ACD are similar; the hypotenuses are in the ratio b/c, so the vertical leg DE = a(b/c) and the horizontal leg AE = b(b/c). Likewise we find BC = b(a/c) (as it should be) and AB = a(a/c). But then BE = CD implies a(a/c)+b(b/c) = c or, multiplying though by ca^2+b^2=c^c.

(Right away you know this proof fails in non Euclidean geometry, because in the first steps it relies on there being one and only one line parallel to CD through A.)


One thought on “Easy foorp

  1. this proof fails in non Euclidean geometry

    Not just that; this even works as a proof by contradiction that there are no non-identical congruent triangles in non-Euclidean geometries. (Probably many proofs of PT can be adapted for that, or for some other result of non-Euclidean geometry, I suppose.)

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