And because the straight line AC is bisected at F, and produced to D, the rectangle AD, DC, together with the square on FC, is equal to the square on FD. [II. 6.
To each of these equals add the square on FE.
Therefore the rectangle AD, DC together with the squares on CF, FE, is equal to the squares on DF, FE. [Axiom 2.
But the squares on CF, FE are equal to the square on CE, because CFE is a right angle; [I. 47.
and the squares on DF, FE are equal to the square on DE.
Therefore the rectangle AD, DC, together with the square on CE, is equal to the square on DE.
But CE is equal to BE;
therefore the rectangle AD, DC, together with the square on BE, is equal to the square on DE.
But the square on DE is equal to the squares on DB, BE, because EBD is a right angle. [I. 47.
Therefore the rectangle AD, DC, together with the square on BE, is equal to the squares on DB, BE.
Take away the common square on BE;
then the remaining rectangle AD, DC is equal to the square on DB. [Axiom 3.
Wherefore, if from any point &c. q.e.d.
Corollary. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circles are equal to one another; namely, the rectangle BA, AE is equal to the rectangle CA, AF; for each of them is equal to the square on the straight line AD, which touches the circle.
