Construction of Circle from Segment

Theorem
From a given segment of a circle, it is possible to construct the complete circle of which it is a segment.

Euclid's Construction
Let $ABC$ be the given segment of a circle whose base is $AC$.

Bisect $AC$ at $D$, draw $DB$ perpendicular to $AC$, and join $AB$.

First suppose that $ABC$ is such that $\angle ABD > \angle BAD$:


 * Euclid-III-25a.png

On $BA$, construct $\angle BAE$ equal to $\angle ABD$.

Join $BD$ through to $E$ and join $EC$.

Then $E$ is the center of the required circle.

Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$:


 * Euclid-III-25b.png

Then $D$ is the center of the required circle.

Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$:


 * Euclid-III-25c.png

On $BA$, construct $\angle BAE$ equal to $\angle ABD$.

The point $E$, which falls on $BD$, is the center of the required circle.

Euclid's Proof
First suppose that $ABC$ is such that $\angle ABD > \angle BAD$:


 * Euclid-III-25a.png

Since $\angle ABE = \angle BAE$, from Triangle with Two Equal Angles is Isosceles we have that $EB = EA$.

Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a right angle, by Triangle Side-Angle-Side Equality we have that $\triangle ADE = \triangle CDE$.

Hence $AE = CE$ both of which are equal to $BE$ from above.

So from Condition for Point to be Center of Circle $E$ is the center of the required circle.

Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$:


 * Euclid-III-25b.png

From Triangle with Two Equal Angles is Isosceles we have that $AD = DB$ and so also equal to $DC$.

So from Condition for Point to be Center of Circle $D$ is the center of the required circle.

Incidentally, note that in this case segment $ABC$ is actually a semicircle.

Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$:


 * Euclid-III-25c.png

The same proof applies:

Since $\angle ABE = \angle BAE$, from Triangle with Two Equal Angles is Isosceles we have that $EB = EA$.

Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a right angle, by Triangle Side-Angle-Side Equality we have that $\triangle ADE = \triangle CDE$.

Hence $AE = CE$ both of which are equal to $BE$ from above.

So from Condition for Point to be Center of Circle $E$ is the center of the required circle.

Alternative Construction and Proof

 * Euclid-III-25d.png

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.