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Proposition 9

Fri, 2007-03-02 14:38 — ashley

If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the point taken is the centre of the circle.

Let $ ABC $ be a circle and $ D $ a point within it, and from $ D $ let more than two equal straight lines, namely $ DA $, $ DB $, $ DC $, fall on the circle $ ABC $;

I say that the point $ D $ is the centre of the circle $ ABC $.

For let $ AB $, $ BC $ be joined and bisected at the points $ E $, $ F $, and let $ ED $, $ FD $ be joined and drawn through to the points $ G $, $ K $, $ H $, $ L $.

Then, since $ AE $ is equal to $ EB $, and $ ED $ is common,

the two sides $ AE $, $ ED $ are equal to the two sides $ BE $, $ ED $;

and the base $ DA $ is equal to the base $ DB $;

therefore the angle $ AED $ is equal to the angle $ BED $. [Prop. 1.8]

Therefore each of the angles $ AED $, $ BED $ is right; [Def. 1.10]

therefore $ GK $ cuts $ AB $ into two equal parts and at right angles.

And since, 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, [Prop. 3.1, Porism]

the centre of the circle is on $ GK $.

For the same reason

the centre of the circle $ ABC $ is also on $ HL $.

And the straight lines $ GK $, $ HL $ have on other point common but the point $ D $;

therefore the point $ D $ is the centre of the circle $ ABC $.

Therefore etc.

Q.E.D.