Condition for Point to be Center of Circle

Theorem
If a point be taken in a circle such that more than two lines fall from that point onto the circle, that point is the center of the circle.

Geometric Proof
Let $$ABC$$ be a circle, and $$D$$ a point in it such that $$DA = DB = DC$$.

Then $$D$$ can be proved to be the center of circle $$ABC$$ as follows.


 * Euclid-III-9.png

Join $$AB$$ and $$BC$$, and bisect them at $$E$$ and $$F$$.

Join $$ED$$ and $$FD$$, and produce them in both directions to $$GK$$ and $$HL$$.

We have that $$AE = EB$$, $$AD = BD$$ and $$ED$$ is common.

From Triangle Side-Side-Side Equality, $$\triangle AED = \triangle BED$$ and so $$\angle AED = \angle BED$$.

So from Book I Definition 10, each of $$\angle AED$$ and $$\angle BED$$ is a right angle.

Hence from the porism to Finding Center of Circle, the center of circle $$ABC$$ is on $$GK$$.

The same applies to $$\triangle BFD = \triangle CFD$$, and so by the same construction, the center of circle $$ABC$$ is also on $$HL$$.

The straight lines $$GK$$ and $$HL$$ have only point $$D$$ in common.

Hence the result.