Kitap 6
29
Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24]. Q. E. F.
PROPOSITION 30.
30
To cut a given finite straight line in extreme and mean ratio.
30
Let AB be the given finite straight line; thus it is required to cut AB in extreme and mean ratio.
30
On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29]
30
Now BC is a square; therefore AD is also a square.
30
And, since BC is equal to CD, let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD.
30
But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, as FE is to ED, so is AE to EB.
30
But FE is equal to AB, and ED to AE.
30
Therefore, as BA is to AE, so is AE to EB.
30
And AB is greater than AE; therefore AE is also greater than EB.
30
Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE. Q. E. F.
PROPOSITION 31.
31
In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.
31
Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.
31
Let AD be drawn perpendicular.
31
Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8]
31
And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1]
