joining straight line will cut the circumference of the given segment at a point which solves the problem.
The given straight line must exceed AB and it must not exceed a certain straight line which we will now determine. Suppose the circumference of the given segment bisected at E: join AE, and produce it to meet the circumference of the described segment at F. Then AE is equal to EB (III. 28), and EB is equal to EF for the same reason that CB is equal to CD. Thus EA, EB, EF are all equal; and therefore E is the centre of the circle of which ADB is a segment (III. 9). Hence AF is the longest straight line which can be drawn from A to the circumference of the described segment; so that the given straight line must not exceed twice AE,
42.To describe an isosceles triangle having each of the angles at the base double of the third angle.
This problem is solved in IV. 10; we may suppose the solution to have been discovered by such an analysis as the following.
Suppose the triangle ABD such a triangle as is required, so that each of the angles at B and D is double of the angle at A.
Bisect the angle at D by the straight line DC. Then the angle ADC is equal to the angle at A; therefore CA is equal to CD. The angle CBD is equal to the angle ADB, by hypothesis; the angle CDB is equal to the angle at A; therefore the third angle BCD is equal to the third angle ABD (I. 32). Therefore BD is equal to CD (I. 6); and therefore BD is equal to AC.
Since the angle BDC is equal to the angle at A, the straight line BD will touch at D the circle described round the triangle ACD {Note on III. 32). Therefore the rectangle AB, BC is equal to the square on BD (III. 36). Therefore the rectangle AB,BC is equal to the square on a AC
Therefore AB is divided at C in the manner required in II. 11.
Hence the synthetical solution of the problem is evident.
