Let the two straight lines AC, BD cut one another at the point E, within the circle ABCD: the rectangle contained by AE,EC shall be equal to the rectangle contained by BE, ED.
If AC and BD both pass through the centre, so that E is the centre, it is evident, since EA, EB, EC, ED are all equal, that the rectangle AE, EC is equal to the rectangle BE,ED.
But let one of them, BD, pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, at the point E. Then, if BD be bisected at F, F is the centre of the circle ABCD; join AF.
Then, because the straight line BD which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles at the point E, [Hypothesis.
AE is equal to EC. [III. 3.
And because the straight line BD is divided into two equal parts at the point F, and into two unequal parts at the point E, the rectangle BE, ED, together with the square on EF, is equal to the square on FB, [II. 5.
that is, to the square on AF.
But the square on AF is equal to the squares on AE,EF.[l.11. Therefore the rectangle BE, ED, together with the square on EF, is equal to the squares on AE, EF. [Axiom 1.
Take away the common square on EF, then the remaining rectangle BE, ED, is equal to the remaining square on AE, that is, to the rectangle AE, EC.
Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, at the point E, but not at right angles. Then, if BD be bisected at F, F is the centre of the circle ABCD; join AF, and from F draw FG perpendicular to AC. [I. 12.
