If on the diameter of a circle a point is taken which is not the center of the circle, and from the point straight lines fall upon the circle, then that is greatest on which passes through the center, the remainder of the same diameter is least, and of the rest the nearer to the straight line through the center is always greater than the more remote; and only two equal straight lines fall from the point on the circle, one on each side of the least straight line.

Let *ABCD* be a circle, and let *AD* be a diameter of it. Let *F* be a point *F* on *AD* which is not the center of the circle. Let *E* be the center of the circle. Let straight lines *FB, FC,* and *FG* fall upon the circle *ABCD* from *F.*

I say that *FA* is greatest, *FD* is least, and of the rest *FB* is greater than *FC,* and *FC* greater than *FG.*

Join *BE, CE,* and *GE.*

Then, since in any triangle the sum of any two sides is greater than the remaining one, the sum of *EB* and *EF* is greater than *BF.*

But *AE* equals *BE,* therefore *AF* is greater than *BF.*

Again, since *BE* equals *CE,* and *FE* is common, the two sides *BE* and *EF* equal the two sides *CE* and *EF.* But the angle *BEF* is also greater than the angle CEF, therefore the base *BF* is greater than the base *CF.*

For the same reason *CF* is also greater than *GF.*

Again, since the sum of *GF* and *FE* is greater than *EG,* and *EG* equals *ED,* the sum of *GF* and *FE* is greater than *ED.*

Subtract *EF* from each. Therefore the remainder *GF* is greater than the remainder *FD.*

Therefore *FA* is greatest, *FD* is least, *FB* is greater than *FC,* and *FC* greater than *FG.*

I say also that from the point *F* only two equal straight lines fall on the circle *ABCD,* one on each side of the least *FD.*

Construct the angle *FEH* equal to the angle *GEF* on the straight line *EF* and at the point *E* on it. Join *FH.*

Then, since *GE* equals *EH,* and *EF* is common, the two sides *GE* and *EF* equal the two sides *HE* and *EF,* and the angle *GEF* equals the angle *HEF,* therefore the base *FG* equals the base *FH.*

I say again that another straight line equal to *FG* does not fall on the circle from the point *F.*

For, if possible, let *FK* so fall.

Above

Then, since *FK* equals *FG,* and *FH* equals *FG, FK* also equals *FH,* the nearer to the straight line through the center being thus equal to the more remote, which is impossible.

Therefore another straight line equal to *GF* does not fall from the point *F* upon the circle. Therefore only one straight line so falls.

Therefore *if on the diameter of a circle a point is taken which is not the center of the circle, and from the point straight lines fall upon the circle, then that is greatest on which passes through the center, the remainder of the same diameter is least, and of the rest the nearer to the straight line through the center is always greater than the more remote; and only two equal straight lines fall from the point on the circle, one on each side of the least straight line.*

Q.E.D.

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