THE PROOF: Let ABC be the triangle and BAC the right angle.

The square BDEC is constructed on BC, the square GB on BA and the square HC on AC. Draw AL parallel to BD or CE and also draw lines AD and FC. I 46; I 31; Post. 1;

Angles BAC and BAG at A are right angles I Def 22;

and each angle on each side of segment BA are adjacent angles which make two right angles when added together; therefore CA and AG are found on the same segment I 14;

Likewise, BA is found on the same segment as AH

Angle DBC is equal to FBA as they are both right angles. If each one is added to angle ABC both angles are still equal.

Therefore DBA is equal to FBC I Def 22; Post 4; CN 2 ;

Since DB is equal to BC and FB is equal to BA, sides AB and BD are equal to sides FB and BC respectively; and as  angle ABD is equal to angle FBC, both triangles are equal. I Def 22; I 4;

Furthermore, the area of the parallelogram BL is double the triangle ABD as it has the same base and is contained by the same parallels BD and AL. The area of square GB is double the triangle FBC as they have the same base and are contained by parallel sides. I 41;

Likewise, parallelogram CL is equal to square HC.

Since square BDEC is constructed on BC, square GB on side BA and square HC on AC, then the square on side BC is equal to the sum of the square on GB and HC. CN 2;

Furthermore, the square BDEC is constructed on BC and the squares GB and HC on BA and AC. Therefore, the square on BC is equal to the sum of the squares on BA and AC.

In this proposition Euclid proves the converse of Pythagoras' theorem.

THE PROOF:

Take the square on BC to be equal to the sum of the squares on AC and AB.

Draw AD from point A at a right angle to side AC, make AD equal to BA and join D to C. I 11; I 3 and Post 1;

Since DA is equal to AB, the square on DA is equal to the square on AB

Add the square on AC to both and you will find that the sum of the squares on AC and AD is equal to the sum of the squares on AC and AB. CN 2;

However, the square on DC is equal to the sum of the squares on AC and AD as they form a right angle, and the square on BC is equal to the sum of the squares on AB and AC through hypothesis. Therefore, the square on CD is equal to the square on BC and side BC is equal to CD. I 47;

Since side DA is equal to AB, side AC is a common side, and since side CD is equal to CB, angle BAC is equal to angle DAC. As angle DAC is a right angle then angle BAC is also a right angle.