# Construction of Circle from Segment

## Theorem

In the words of Euclid:

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

## Proof 1

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 $\angle ABD > \angle BAD$:

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 $\angle ABD = \angle BAD$:

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

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

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

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

### Proof of Construction

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

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 $\angle ABD = \angle BAD$:

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 $\angle ABD < \angle BAD$:

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.

$\blacksquare$

## Proof 2

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.

$\blacksquare$