Construction of Circle from Segment/Proof 2

Theorem

In the words of Euclid:

Given a segment of a circle, to describe the complete circle of which it is a segment.

(The Elements: Book $\text{III}$: Proposition $25$)


Proof

Choose any point $C$ on the circumference.

Bisect $AC$ at $D$ and $BC$ at $E$ and construct a perpendicular $DF$ and $EF$ from each through the point of bisection.

The point of intersection $F$ is the center of the required circle.


$AFC$ and $CFB$ are isosceles triangles and so $AF, CF$ and $BF$ are all equal.

The result follows from Condition for Point to be Center of Circle.

$\blacksquare$


Sources

  • 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions
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