T H E E L E M E N T 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, AF is equal to FC. |
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