 | William Desborough Cooley - Geometry - 1840 - 106 pages
...proportionals be numbers, the solution of this Problem may be conveniently derived from the principle that the square of the mean is equal to the rectangle contained by (or the product of) the extremes (vi. Prop. 17). From the square of the given sum of the extremes,... | |
 | Peter Michael Harman, Alan E. Shapiro - Biography & Autobiography - 2002 - 552 pages
...Now because AB = AX and DX is perpendicular to AY, so DA/ AX = AX/ A Y. And in continued proportions, the square of the mean is equal to the rectangle contained by the extremes. Therefore, [AB 2 = AX 2 = DA-A Y]. Corollary. If I had said that the rectangle contained by the segments... | |
| |
And in continued proportions, the square of the mean is equal to the rectangle contained by the extremes. 