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To cut a given finite straight line in extreme and mean ratio.

Let AB be the given finite straight line;

thus it is required to cut AB in extreme and mean ratio.

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]

Now BC is a square;

therefore AD is also a square.

And, since BC is equal to CD, let CE be subtracted from each;

therefore the remainder BF is equal to the remainder AD.

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.

But FE is equal to AB, and ED to AE.

Therefore, as BA is to AE, so is AE to EB.

And AB is greater than AE;

therefore AE is also greater than EB.

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.

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