In a circle equal straight lines are equally distant from the centre, and those which are equally distant from the centre are equal to one another.
Proof
Let AB and CD be chords with AB=CD. From the centre E
drop perpendiculars EF to AB and EG to CD (I.12). By III.3,
AF=FB=AB/2 and CG=GD=CD/2, so AF=CG. Join EA,
EC (both radii, so equal). In right triangles △EAF
and △ECG (right angles at F, G), I.47 gives EA2=AF2+EF2 and EC2=CG2+EG2. Subtracting (Common Notion
3) and using EA=EC, AF=CG gives EF2=EG2, hence EF=EG. Conversely, if EF=EG, the same I.47 identity gives AF=CG and hence AB=CD.