T H E E L E M E N T S |
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; therefore 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, |
||
Previous Page Return to Propositions Next Page |