Kitap 6
25
Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF; therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16]
25
But the triangle ABC is equal to the parallelogram BE; therefore the triangle KGH is also equal to the parallelogram EF.
25
But the parallelogram EF is equal to D; therefore KGH is also equal to D.
25
And KGH is also similar to ABC.
25
Therefore one and the same figure KGH has been constructed similar to the given rectilineal figure ABC and equal to the other given figure D. Q. E. D. to which the figure to be constructed must be similar, literally to which it is required to construct (one) similar, ᾧ δεῖ ὅμοιον συστήσασθαι.
PROPOSITION 26.
26
If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole
26
For from the parallelogram ABCD let there be taken away the parallelogram AF similar and similarly situated to ABCD, and having the angle DAB common with it; I say that ABCD is about the same diameter with AF.
26
For suppose it is not, but, if possible, let AHC be the diameter lt of ABCD gt, let GF be produced and carried through to H, and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31]
26
Since, then, ABCD is about the same diameter with KG, therefore, as DA is to AB, so is GA to AK. [VI. 24]
26
But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE; therefore also, as GA is to AK, so is GA to AE. [V. 11]
26
Therefore GA has the same ratio to each of the straight lines AK, AE.
26
Therefore AE is equal to AK [V. 9], the less to the greater : which is impossible.
26
Therefore ABCD cannot but be about the same diameter with AF; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF.
26
Therefore etc. Q. E. D.
PROPOSITION 27.
