Let the parallelogram AC be equiangular to the parallelogram CF, having the angle BCD equal to the angle ECG: the parallelogram AC shall have to the parallelogram CF the ratio which is compounded of the ratios of their sides.
Let BC and CG be placed in a straight line;
therefore DC and CE are also in a straight line; [I. 1 4,
complete the parallelogram DG;
take any straight line K, and make K to L as BC is to CG and L to M as DC to CE; [VI 12.
then the ratios of K to L and of L to M are the same with the ratios of the sides, namely, of BC to CG and of DC to CE.
But the ratio of K to M is that which is said to be compounded of the ratios of K to L and of L to M; [V. Def. A.
therefore K has to M the ratio which is compounded of the ratios of the sides.
Now the parallelogram AC is to the parallelogram CH as BC is to CG; [VI. 1.
but BC is to CG as ^is to Z; [Construction.
therefore the parallelogram AC is to the parallelogram CH as Kis to L. [V. 11.</br Again, the parallelogram CH is to the parallelogram CF as DC is to CE; [VI 1.
but DC is to CE as L is to M; [Construction.
therefore the parallelogram CH is to the parallelogram CF as L is to M. [V. 11.
Then, since it has been shewn that the parallelogram AC is to the parallelogram CH as K is to L,
and that the parallelogram CH is to the parallelogram CF as L is to M,
therefore, ex sequali, the parallelogram AC is to the parallelogram CF as K is to M. [V. 22. But K has to M the ratio which is compounded of the ratios of the sides;
