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Proposition 1

Fri, 2007-01-26 18:41 — ashley

To find the centre of a given circle.

Let $ ABC $ be the given circle;

thus it is required to find the centre of the circle $ ABC $.

Let a straight line $ AB $ be drawn through it at random, and let it be bisected at the point $ D $;

from $ D $ let $ DC $ be drawn at right angles to $ AB $ and let it be drawn through to $ E $;

let $ CE $ be bisected at $ F $;

I say that $ F $ is the centre of the circle $ ABC $.

For suppose it is not, but, if possible, let $ G $ be the centre,

and let $ GA $, $ GD $, $ GB $ be joined.

Then, since $ AD $ is equal to $ DB $,

and $ DG $ is common,

the two sides $ AD $, $ DG $ are equal to the two sides $ BD $, $ DG $ respectively;

and the base $ GA $ is equal to the base $ GB $ for they are radii;

therefore the angle $ ADG $ is equal to the angle $ GDB $. [Prop. 1.8]

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [Def. 1.10]

therefore the angle $ GDB $ is right.

But the angle $ FDB $ is also right;

therefore the angle $ FDB $ is equal to the angle $ GDB $, the greater to the less; which is impossible.

Therefore $ G $ is not the centre of the circle $ ABC $.

Similarly we can prove that neither is any other point except $ F $.

Therefore the point $ F $ is the centre of the circle $ ABC $.

[PORISM. From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line.]

Q.E.F.