Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/234

210 prallelogram EG; let GF meet AHC at H, and through H draw HK parallel to AD or BC. [1. 31.

Then the parallelograms ABCD and AKHG are about the same diameter, and are therefore similar to one another; [VI. 24. therefore DA is to AB as GA is to AK. But because ABCD and AEFG are similar parallelograms, [Hypothesis. therefore DA is to AB as GA is to AE. [VI. Definition 1. Therefore GA is to AK as GA is to AE, [V. 11. that is, GA has the same ratio to each of the straight lines AK and AE, and therefore AK is equal to AE, [V. 9. the less to the greater; which is impossible. Therefore the parallelograms ABCD and AEFG must have their diameters in the same straight line, that is, they are about the same diameter.

Wherefore, if two similar parallelograms &c.

PROPOSITION 30. PROBLEM. To cut a given straight line in extreme and mean ratio.

Let AB be the given straight line: it is required to cut it in extreme and mean ratio.

Divide AB at the point C, so that the rectangle contained by AB, BC may be equal to the square on AC. [II.11. Then, because the rectangle AB, BC is equal to the square on AC, [Construction. therefore AB is to AC a,s AC is to CB. [VI. 17.

Wherefore AB is cut in extreme and mean ratio at the point C. [VI. Definition 3.