of it, make the adjacent angles ACE,ACB together equal to two right angles,
therefore BC and CE are in one straight line. [I. 14.
Wherefore, if two triangles &c. q.e.d.
PROPOSITION 33. THEOREM.
In equal circles, angles, whether at the centres or at the circumferences, have the same ratio which the arcs on which they stand have to one another; so also have the sectors.
Let 'ABC and BEF be equal circles, and let BGC and EHF be angles at their centres, and BAC and EDF angles at their circumferences: as the arc BC is to the arc EF so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF; and so also shall the sector BGC be to the sector EHF.
Take any number of arcs CK, KL, each equal to BC, and also any number of arcs FM, MN each equal to EF; and join GK, GL, HM, HN.
Then, because the arcs BC, CK, KL, are all equal, [Constr.
the angles BCG, CGK, KGL are also all equal; [III. 27.
and therefore whatever multiple the arc BL is of the arc BC, the same multiple is the angle BGL of the angle BGC.
For the same reason, whatever multiple the arc EN is of the arc EF, the same multiple is the angle EHN of the angle EHF.
