| BAC : EDF | :: BOC : EPF | [V. xv.]; | |||||||||||||||||||||
| but | BOC : EPF | :: BC : EF. | |||||||||||||||||||||
| Hence | BAC : EDF | :: BC : EF | [V. xi.]. |
2. The sector BOC : sector EPF :: BC : EF.
Dem.—The same construction being made, since the arc BC is equal to CG, the angle BOC is equal to COG. Hence the sectors BOC, COG are congruent (see Observation, Proposition xxix., Book III.); therefore they are equal. In like manner the sectors COG, GOH are equal. Hence there are as many equal sectors as there are equal arcs; therefore the arc BH and the sector BOH are equimultiples of the arc BC and the sector BOC. In the same manner it may be proved that the arc EJ and the sector EPJ are equimultiples of the arc EF and the sector EPF; and it is evident, by superposition, that if the arc BH is greater than, equal to, or less than the arc EJ, the sector BOH is greater than, equal to, or less than the sector EPJ. Hence [V. Def. v.] the arc BC : EF :: sector BOC : sector EPF.
The second part may be proved as follows:—
Sector BOC =
rectangle contained by the arc BC, and the radius of the circle ABC [xx. Ex. 14] and sector EPF =
rectangle contained by the arc EF and the radius of the circle EDF; and since the circles are equal, their radii are equal. Hence, sector BOC : sector EPF :: arc BC : arc EF.
Questions for Examination on Book VI.