from the centre, when the perpendiculars drawn to them from the centre are equal; [III. Definition 4.
therefore AB, CD are equally distant from the centre.
Next, let the straight lines AB, CD be equally distant from the centre, that is, let EF be equal to EG, AB shall be equal to CD.
For, the same construction being made, it may be shewn, as before, that AB is double of AF, and CD double of CG and that the squares on EF,FA are equal to the squares on EG,GC; but the square on EF is equal to the square on EG, because EF is equal to EG; [Hypothesis. therefore the remaining square on FA is equal to the remaining square on GC, [Axiom 3.
and therefore the straight line AF is equal to the straight line CG.
But AB was shewn to be double of AF, and CD double of CG. Therefore AB is equal to CD. [Axiom 6.
Wherefore, equal straight lines &c. q.e.d.
PROPOSITION 15. THEOREM.
The diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is cdways greater than one more remote; and the greater is nearer to the centre than the less.
Let ABCD be a circle, of which AD is a diameter, and E the centre; and let BC be nearer to the centre than FG:
AD shall be greater than any straight line BC which is not a diameter, and BC shall be greater than FG.
From the centre E draw EH, EK perpendiculars to BC, FG, [I. 12.
and join EB, EC, EF.
Then, because AE is equal to BE, and ED to EC, [I.Def.15.
therefore AD is equal to BE, EC; [Axiom 2.
