Book 4
1
Then, if BC is equal to D, that which was enjoined will have been done; for BC has been fitted into the circle ABC equal to the straight line D.
1
But, if BC is greater than D, let CE be made equal to D, and with centre C and distance CE let the circle EAF be described; let CA be joined.
1
Then, since the point C is the centre of the circle EAF, CA is equal to CE.
1
But CE is equal to D; therefore D is also equal to CA.
1
Therefore into the given circle ABC there has been fitted CA equal to the given straight line D.
PROPOSITION 2.
2
In a given circle to inscribe a triangle equiangular with a given triangle.
2
Let ABC be the given circle, and DEF the given triangle; thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF.
2
Let GH be drawn touching the circle ABC at A [III. 16, Por.]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined.
2
Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32]
2
But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF.
