|Book III Proposition 33|
Then, since AF is again 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 so pass, as ABE.
Now, since AD is drawn at right angles to the diameter AE from its extremity,
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