BED is equal to the angle AEC by hypothesis; and the angle AEC is equal to the angle DEG (I. 15). Hence the
triangles BEF and GEF are equal in all respects (I. 26); therefore FG is equal to FB.
This result shews how we may synthetically solve the problem. Draw BF perpendicular to CD, and produce it to G, so that FG may be equal to FB; then join AG, and AG will intersect CD at the required point.
40. To divide a given straight line into two parts such that the difference of the squares on the parts may be equal to a given square.
Let AB he the given straight line, and suppose C the required point.
Then the difference of the squares on C and BG is to be equal to a given square. But the difference of the squares on AC and BC is equal to the rectangle contained by their sum and difference; therefore this rectangle must be equal to the given square. Hence we have the following synthetical solution. On AB describe a rectangle equal to the given square (I. 45); then the difference of AC and CB will be equal to the side of the rectangle adjacent to AB, and is therefore known. And the sum of AC and CB is known. Thus AC and CB are known.
It is obvious that the given square must not exceed the square on AB, in order that the problem may be possible.
There are two positions of C, if it is not specified which of the two segments AC and CB is to be greater than the other; but only one position, if it is specified.
