If I were allowed a small philosophical leeway, I'd argue that two correct proofs are always the same. For sure they may contain different words or make use of different "abstractions", but it just seems to me that these abstractions should be equivalent if one were willing to unravel it all to a certain degree. Essentially, all proof is, is a statement that says "this is true" and no matter which language you use to…
1. Prove that the interior angles of a triangle sum to 180 degrees.
First proof: draw a line parallel to one of the triangle's sides passing through its opposite vertex. There are three angles on one side of this line, and they obviously add to 180 degrees because it's a line. One of the three angles is directly one of the triangle's interior angles; the other two can be shown to be equal to the triangle's other two interior angles. (Try drawing it out.)
Second proof: start at one side of the triangle and walk around it. By the time you return to where you started, you must have turned 360 degrees. Thus the sum of the exterior angles is 360 degrees. Each interior angle is 180 minus the corresponding exterior angle, and there are three of them, so calling the interior angles A, B, C and the exterior angles A', B', C' we have A'+B'+C' = 360 implies (180-A) + (180-B) + (180-C) = 360 implies 540 - A - B - C = 360 implies 180 = A + B + C.
2. Prove that the sum of the first N numbers is N(N+1)/2.
First proof: sum the first and last number to get 1 + N, then the second and second-to-last to get 2 + (N-1) = 1 + N, repeating until you get to the middle. There are N/2 such pairs, giving a total of (1 + N)N/2. (This assumed that there were an even number of terms; consider the odd case too.)
Second proof: proceed by induction. For the base case, it's true for N=1 because 1*2/2 = 1. For the inductive case, suppose it's true for N-1. Then 1 + 2 + ... + N-1 + N = (1 + 2 + ... + N-1) + N = N(N-1)/2 + N = N(N-1)/2 + 2N/2 = N(N+1)/2.