Kitap 6
8
For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC.
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]
