Book 6
8
Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]
8
Therefore etc.
8
Porism. From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Q. E. D.
PROPOSITION 9.
9
From a given straight line to cut off a prescribed part.
9
Let AB be the given straight line; thus it is required to cut off from AB a prescribed part.
9
Let the third part be that prescribed.
9
Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3]
9
Let BC be joined, and through D let DF be drawn parallel to it. [I. 31]
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.
