A proof of Pythagoras Theorem.

© Stu Savory, 2005.

A proof of Pythagoras Theorem.

Draw a square with sides of length A+B. Mark off the length B from each corner going around in a clockwise direction (say). Join up these marks to make a smaller, tilted square inside the larger square. Label the sides of the smaller, tilted square C.

Now the area of the larger square is (A+B)2, as is obvious to see.

The area of each of the triangles is (A*B)/2, as can be seen in the triangle on the left. So the total of all 4 triangles in the larger square is 4*(A*B)/2 = 2*(A*B). The area of the tilted square is C2, obviously.

Therefore (A+B)2 = 2*(A*B) + C2.

Expanding the left side we get A2 + 2*(A*B) + B2 = 2*(A*B) + C2.
Cancelling 2*(A*B) from both sides, we get A2 + B2 = C2.

Quod erat demonstrandum.

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