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S		 Book III   Proposition 36
Next let DCA not be through the center of the circle ABC;
    let the center E be taken;
and from E
    let EF be drawn perpendicular AC;
    let EB, EC, ED be joined.
Then the angle EBD is right. [III.18]

And, since a straight line EF throught the center cuts a straight line AC not through the center at right angles,
it also bisects it; [III.3]
therefore
    AF is equal to FC.
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