Equal straight lines in a circle are equally distant from the center, and those which are equally distant from the center equal one another.

Let *AB* and *CD* be equal straight lines in a circle *ABDC.*

I say that *AB* and *CD* are equally distant from the center.

Take the center *E* of the circle *ABDC.* Draw *EF* and *EG* from *E* perpendicular to *AB* and *CD,* and join *AE* and *EC.*

Then, since a straight line *EF* passing through the center cuts a straight line *AB* not passing through the center at right angles, it also bisects it. Therefore *AF* equals *FB.* Therefore *AB* is double *AF.*

For the same reason *CD* is also double *CG.* But *AB* equals *CD,* therefore *AF* also equals *CG.*

Also, since *AE* equals *EC,* the square on *AE* also equals the square on *EC.* But the sum of the squares on *AF* and *EF* equals the square on *AE,* for the angle at *F* is right, and the sum of the squares on *EG* and *GC* equals the square on *EC,* for the angle at *G* is right. Therefore the sum of the squares on *AF* and *FE* equals the sum of the squares on *CG* and *GE,* of which the square on *AF* equals the square on *CG,* for *AF* equals *CG.* Therefore the remaining square on *FE* equals the square on *EG.* Therefore *EF* equals *EG.*

But straight lines in a circle are said to be equally distant from the center when the perpendiculars drawn to them from the center are equal. Therefore *AB* and *CD* are equally distant from the center.

Next, let the straight lines *AB* and *CD* be equally distant from the center, that is, let *EF* equal *EG.*

I say that *AB* also equals *CD.*

For, with the same construction, we can prove, as before, that *AB* is double *AF,* and *CD* double *CG.* And, since *AE* equals *CE,* the square on *AE* equals the square on *CE.* But the sum of the squares on *EF* and *FA* equals the square on *AE,* and the sum of the squares on *EG* and *GC* equals the square on *CE.*

Therefore the sum of the squares on *EF* and *FA* equals the sum of the squares on *EG* and *GC,* of which the square on *EF* equals the square on *EG,* for *EF* equals *EG.* Therefore the remaining square on *AF* equals the square on *CG.* Therefore *AF* equals *CG.* And *AB* is double *AF,* and *CD* double *CG,* therefore *AB* equals *CD.*

Therefore *equal straight lines in a circle are equally distant from the center, and those which are equally distant from the center equal one another.*

Q.E.D.

This proposition is used in the next one.