Book 2
8
But the gnomon STU and OH are the whole square AEFD, which is described on AD. therefore four times the rectangle AB, BD together with the square on AC is equal to the square on AD
8
But BD is equal to BC; therefore four times the rectangle contained by AB, BC together with the square on AC is equal to the square on AD, that is to the square described on AB and BC as on one straight line.
8
Therefore etc. Q. E. D.
Proposition 9.
9
If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.
9
For let a straight line AB be cut into equal segments at C, and into unequal segments at D;
9
I say that the squares on AD, DB are double of the squares on AC, CD.
9
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.
9
Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC.
9
And, since the angle at C is right, the remaining angles EAC, AEC are equal to one right angle. [I. 32]
9
And they are equal; therefore each of the angles CEA, CAE is half a right angle.
9
For the same reason each of the angles CEB, EBC is also half a right angle; therefore the whole angle AEB is right.
9
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]
