Another proof of Pythagoras' theorem from Tutorial 4

Back to Geometry homepage

In the tutorials we have seen a number of different proofs of Pythagoras' theorem. This page contains some more explanation of the proof given in Tutorial 4.

The diagram below shows a right-angled triangle \(\Delta ABC\). Pythagoras' theorem states that the area of the square on the hypotenuse \(AC\) (defined by the red lines in the diagram) equals the sum of the squares on the sides \(AB\) (coloured blue) and \(BC\) (coloured brown). Equivalently, \(|AC|^2 = |AB|^2 + |BC|^2\).

Beginning with the blue and brown square containing the two green triangles, subtract the area of the two green triangles and add the area of the two red triangles. The result is the square defined by the red lines, with one side equal to \(AC\). Since all of the triangles are congruent, then we have added and subtracted the same area. Therefore the area of the large square is the sum of the areas of the blue and brown squares, which is exactly the statement of Pythagoras' theorem.