Let BC be joined, and through D let DF be drawn parallel to it. [[I. 31](elem.1.31)]

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](elem.6.2)]

But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF.

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.