Construction of Circle from Segment/Proof 2
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Theorem
In the words of Euclid:
(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