Book 2
8
But MQ is equal to QL, for they are complements of the parallelogram ML; [I. 43] therefore AG is also equal to RF; therefore the four areas AG, MQ, QL, RF are equal to one another. Therefore the four are quadruple of AG. But the four areas CK, KD, GR, RN were proved to be quadruple of CK; therefore the eight areas, which contain the gnomon STU, are quadruple of AK.
8
Now, since AK is the rectangle AB, BD, for BK is equal to BD, therefore four times the rectangle AB, BD is quadruple of AK.
8
But the gnomon STU was also proved to be quadruple of AK; therefore four times the rectangle AB, BD is equal to the gnomon STU.
8
Let OH, which is equal to the square on AC, be added to each; therefore four times the rectangle AB, BD together with the square on AC is equal to the gnomon STU and OH.
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.
