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Ancient Proof of Pythagoras Theorem

Pythagoras never gave any proof to the famous theorem named after him. And I have read that Pythagoras actually had been to India to study geometry from where he picked up the statement of this theorem, and this theorem was well known to Indians centuries before Pythagoras, and the Indian mathematician Baudhayana was the one who actually discovered this theorem.

Now my question is, is there any ancient proof of the Pythagoras theorem which was published before the times of Pythagoras. I want mathematical details of such a mathematical proof.

Ancient Proof of Pythagoras Theorem

Ancient Proof of the Pythagoras Theorem

ABCD is a square with the length of each side being (a+b). Hence,

Area of Square ABCD = (a+b)2 ( eq. 1 )

Now consider another square inscribed inside ABCD i.e. EFGH whose length of the sides is c. (Look at the above diagram). Also there are 4 right angled triangles (every angle in a square is a right angle) inside the square ABCD. We also see that the area of all these 4 right angled triangles is equal and is given by
1/2 * b * a.

Area of each right angled triangle = (1/2) * a * b ( eq. 2 )
We can also see that,

Area of the Square EFGH = c2 ( eq. 3)

Now we see that the Area of the Square ABCD is equal to the Sum of the Areas of the 4 right angled triangles and the Area of the square EFGH. i.e.

Area of Square ABCD = (4 x Area of a right angled trianlge) + Area of Square EFGH
Substituting equations 1,2 and 3 in the above equation we get

=> (a+b)2 = [4 x (1/2) x a x b] + [ c2 ]
=> (a+b)2 = 2ab + c2
=> a2 + b2 + 2ab = 2ab + c2
=> a2 + b2 = c2

This is nothing but the theorem what we know today in the name of Pythagoras! This proof clearly shows that in a right angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides.

This page was last modified on November 30, 2012 at 3:54 pm