In equal circles straight lines that cut off equal circumferences are equal.

Let *ABC* and *DEF* be equal circles, and in them let equal circumferences *BGC* and *EHF* be cut off. Join the straight lines *BC* and *EF.*

I say that *BC* equals *EF.*

Take the centers *K* and *L* of the circles. Join *BK, KC, EL,* and *LF.*

Now, since the circumference *BGC* equals the circumference *EHF,* the angle *BKC* also equals the angle *ELF.*

And, since the circles *ABC* and *DEF* are equal, the radii are also equal, therefore the two sides *BK* and *KC* equal the two sides *EL* and *LF,* and they contain equal angles, therefore the base *BC* equals the base *EF.*

Therefore *in equal circles straight lines that cut off equal circumferences are equal.*

Q.E.D.