Let triangle ABC be a right-angled triangle having the angle BAC right.

I say that the square on BC equals the sum of the square on AB and square on AC.

Describe the square BDEC on line BC, and the square GB and HC on line BA and line AC. Draw line AL through point A parallel to either line BD or line CE, and join AD and FC.

Since each of the angles BAC and BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines AC and AG not lying on the same side make the adjacent angles equal to two right angles, therefore line AC is in a straight line with line AG.

For the same reason line AB is also in a straight line with line AH.

Since the angle DBC equals the angle FBA, for each is right, add the angle ABC to each, therefore the whole angle DBA equals the whole angle FBC.

Since line BD equals line BC, and line BF equals line AB, the two sides AB and BD equal the two sides line BF and line BC respectively, and the angle ABD equals the angle FBC, therefore the base AD equals the base FC, and the triangle ABD equals the triangle BCF.

Now the parallelogram BL is double the triangle ABD, for they have the same base line BD and are in the same parallels line BD and line AL. And the square GB is double the triangle BCF, for they again have the same base FB and are in the same parallels line BF and line CG.

Therefore the parallelogram BL also equals the square GB.

Similarly, if line AE and line BK are joined, the parallelogram CL can also be proved equal to the square HC. Therefore the whole square BDEC equals the sum of the two squares GB and HC.

And the square BDEC is described on line BC, and the squares GB and HC on line BA and line AC.

Therefore the square on line BC equals the sum of the squares on line BA and line AC.

Therefore in right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.

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