Construction of Circle from Segment/Proof 1

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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

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.


Proof of Construction

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.

$\blacksquare$


Historical Note

This theorem is Proposition $25$ of Book $\text{III}$ of Euclid's The Elements.


Sources