Proving Pythagoras’ Theorem

Hopefully everyone reading this can state Pythagoras’ Theorem by heart – but, just in case:

“Given a right-angled triangle, the square of the length of the hypotenuse is the sum of the squares of the two shorter sides.” 

Stated more simply, and with the aid of a diagram:


c2 = a2b2



But how can this result be proved? It’s no good drawing lots of right-angled triangles and measuring their sides – firstly, because it’s impossible to measure accurately; and secondly, showing that something is true in lots of cases doesn’t prove it is always true.

Believe it or not, there are well over 300 different proofs! Here’s one of the simplest, using the geometry of squares and triangles.

Take a look at this diagram where I have rotated the triangle (above) three times through 90°, and placed each of the four triangles so as each hypotenuse forms an inner square with side length c.

Let’s consider the area of the whole shape and its components.

Area of whole shape
This is (a + b)(a + b) = {a^2} + 2ab + {b^2}

Area of the blue triangles
Each blue triangle has area given by the formula {1 \over 2} \times {\rm{base}} \times {\rm{height}} = {1 \over 2}ab. So the total of the four triangles is 4 \times {1 \over 2}ab = 2ab.

Area of the blue triangles and yellow square combined
So we can now add together the blue and the yellow areas to get 2ab + {c^2}. But this is exactly the same as the area of the whole shape, so we can write:

{a^2} + 2ab + {b^2} = 2ab + {c^2}

By subtracting 2ab from both sides, we end up with {a^2} + {b^2} = {c^2}  q.e.d.

(q.e.d stands for “quod erat demonstrandum”, Latin for “which was to be demonstrated”, and is the standard way to mark the end of a proof).

Mathematicians like picking holes in proofs, and there’s one in the proof above – there’s a statement I made which needs further examination. Did you spot it? When I first drew the diagram, I said that the four hypotenuses formed an inner square. However, just because a quadrilateral has four sides the same length it doesn’t follow that it’s a square – it could be a rhombus. To satisfy ourselves that it is indeed a square, we need to show that each of its angles is 90°. Have a look at the diagram and see if you can see why.

To show each angle of the yellow quadrilateral is 90°
Look at the diagram below which shows just part of the overall shape.

Since the top triangle is right-angled, xy = 90°. But that means z = 90° as well since xy (the one in the bottom triangle) and z all lie on a straight line. This argument can be repeated for each angle of the quadrilateral, thus proving it is a square.

In fact, you don’t need to prove that all four angles are 90° – how many would suffice?


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