|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,
AF is equal to FC.
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