Book 2
9
For let a straight line AB be cut into equal segments at C, and into unequal segments at D;
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I say that the squares on AD, DB are double of the squares on AC, CD.
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For let CE be drawn from C at right angles to AB, and let it be made equal to either AC or CB; let EA, EB be joined, let DF be drawn through D parallel to EC, and FG through F parallel to AB, and let AF be joined.
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Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC.
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And, since the angle at C is right, the remaining angles EAC, AEC are equal to one right angle. [I. 32]
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And they are equal; therefore each of the angles CEA, CAE is half a right angle.
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For the same reason each of the angles CEB, EBC is also half a right angle; therefore the whole angle AEB is right.
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And, since the angle GEF is half a right angle, and the angle EGF is right, for it is equal to the interior and opposite angle ECB, [I. 29] the remaining angle EFG is half a right angle; [I. 32] therefore the angle GEF is equal to the angle EFG, so that the side EG is also equal to GF. [I. 6]
9
Again, since the angle at B is half a right angle, and the angle FDB is right, for it is again equal to the interior and opposite angle ECB, [I. 29] the remaining angle BFD is half a right angle; [I. 32] therefore the angle at B is equal to the angle DFB, so that the side FD is also equal to the side DB. [I. 6]
9
Now, since AC is equal to CE, the square on AC is also equal to the square on CE; therefore the squares on AC, CE are double of the square on AC.
