If in a circle two straight lines cut one another, then the rectangle contained by the segments of the one equals the rectangle contained by the segments of the other.

For in the circle *ABCD* let the two straight lines *AC* and *BD* cut one another at the point *E.*

I say that the rectangle *AE* by *EC* equals the rectangle *DE* by *EB.*

If now *AC* and *BD* are through the center, so that *E* is the center of the circle *ABCD,* it is manifest that, *AE, EC, DE,* and *EB* being equal, the rectangle *AE* by *EC* also equals the rectangle *DE* by *EB.*

Next let *AC* and *DB* not be through the center. Take the center *F* let the center of *ABCD.* Draw *FG* and *FH* from *F* perpendicular to the straight lines *AC* and *DB.* Join *FB, FC,* and *FE.*

Then, since a straight line *GF* through the center cuts a straight line *AC* not through the center at right angles, it also bisects it, therefore *AG* equals *GC.*

Since, then, the straight line *AC* has been cut into equal parts at *G* and into unequal parts at *E,* the rectangle *AE* by *EC* together with the square on *EG* equals the square on *GC.*

Add the square on *GF.* Therefore the rectangle *AE* by *EC* plus the sum of the squares on *GE* and *GF* equals the sum of the squares on *CG* and *GF.*

But the square on *FE* equals the sum of the squares on *EG* and *GF,* and the square on *FC* equals the sum of the squares on *CG* and *GF.* Therefore the rectangle *AE* by *EC* plus the square on *FE* equals the square on *FC.*

And *FC* equals *FB,* therefore the rectangle *AE* by *EC* plus the square on *EF* equals the square on *FB.*

For the same reason, also, the rectangle *DE* by *EB* plus the square on *FE* equals the square on *FB.*

But the rectangle *AE* by *EC* plus the square on *FE* was also proved equal to the square on *FB,* therefore the rectangle *AE* by *EC* plus the square on *FE* equals the rectangle *DE* by *EB* plus the square on *FE.*

Subtract the square on *FE* from each. Therefore the remaining rectangle *AE* by *EC* equals the rectangle *DE* by *EB.*

Therefore *if in a circle two straight lines cut one another, then the rectangle contained by the segments of the one equals the rectangle contained by the segments of the other.*

Q.E.D.

This proposition is not used in the rest of the *Elements.*