|Book III Proposition 33|
Then, since AF is equal to FB, and FG is common,
the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG;
the base AG is equal to the base BG. [I.4]
Therefore the circle described with center G and distance GA will pass through B also.
Let it be drawn, and let it be ABE;
let EB be joined.
Now, since AD is drawn from A, the extremity of the diameter AE, at right angles to AE,
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