Proof: Start with the right triangle ABC with right angle at C. Draw a square on the hypotenuse AB, and translate the original triangle ABC along this square to get a congruent triangle A'B'C' so that its hypotenuse A'B' is the other side of the square (but the triangle A'B'C' lies inside the square). Draw perpendiculars A'E and B'F from the points A' and B' down to the line BC. Draw a line AG to complete the square ACEG.
Note that ACEG is a square on the leg AC of the original triangle. Also, the square EFB'C' has side B'C' which is equal to BC, so it equals a square on the leg BC. Thus, what we need to show is that the square ABB'A' is equal to the sum of the squares ACEG and EFB'C'.
But that's pretty easy by cutting and pasting. Start with the big square ABB'A'. Translate the triangle A'B'C' back across the square to triangle ABC, and translate the triangle AA'G across the square to the congruent triangle BB'F. Paste the pieces back together, and you see you've filled up the squares ACEG and EFB'C'. Therefore, ABB'A' = ACEG + EFB'C', as required.
Note that ACEG is a square on the leg AC of the original triangle. Also, the square EFB'C' has side B'C' which is equal to BC, so it equals a square on the leg BC. Thus, what we need to show is that the square ABB'A' is equal to the sum of the squares ACEG and EFB'C'.
But that's pretty easy by cutting and pasting. Start with the big square ABB'A'. Translate the triangle A'B'C' back across the square to triangle ABC, and translate the triangle AA'G across the square to the congruent triangle BB'F. Paste the pieces back together, and you see you've filled up the squares ACEG and EFB'C'. Therefore, ABB'A' = ACEG + EFB'C', as required.
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