In like manner we may solve the problem, to produce a given straight line so that the square on the whole straight line made up of the given straight line and the part produced, may exceed the square on the part produced by a given square, which is not less than the square on the given straight line.
The two problems may be combined in one enunciation thus, to divide a given straight line internally or externally so that the difference of the squares ow the segments may be equal to a given square.
41. To find a point in the circumference of a given segment of a circle, so that the straight lines which join the point to the extremities of the straight line on which the segment stands may be together equal to a given straight line.
Let ACB be the circumference of the given segment, and suppose C the required point, so that the sum of AC and CB is equal to a given straight line.
Produce AC to D so that CD may be equal to CB; and join DB.
Then AD is equal to the given straight line. And the angle ACB is equal to the sum of the angles CDB and CBD (I. 32), that is, to twice the angle CDB (I. 5). Therefore the angle ADB is half of the angle in the given segment. Hence we have the following synthetical solution. Describe on AB a segment of a circle containing an angle equal to half the angle in the given segment. With A as centre, and a radius equal to the given straight line, describe a circle. Join A with a point of intersection of this circle and the segment which has been described; this
