Kitap 6
9
Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2]
9
But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF.
9
Therefore from the given straight line AB the prescribed third part AF has been cut off. Q. E. F. any angle. The expression here and in the two following propositions is τυχοῦσα γωνία, corresponding exactly to τυχὸν σημεῖον which I have translated as a point (taken) at random ; but an angle (taken) at random would not be so appropriate where it is a question, not of taking any angle at all, but of drawing a straight line casually so as to make any angle with another straight line.
PROPOSITION 10.
10
To cut a given uncut straight line similarly to a given cut straight line.
10
Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31]
10
Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34]
10
Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]
10
But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF.
