Chord Lies Inside its Circle

Theorem

In the words of Euclid:

If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.

Let $ABC$ be a circle where $A, B$ are two points chosen arbitrarily on its circumference.

Aiming for a contradiction, suppose the straight line $AB$ does not lie entirely within the circle $ABC$.

Then it will lie wholly outside (otherwise it will cross the circumference somewhere, and so we can take the part of $AB$ which does lie outside).

Draw $DA$, $DB$ and $DE$, and let $F$ be the point where $DE$ crosses the circumference.

As $DA = DB$ then from Isosceles Triangle has Two Equal Angles it follows that $\angle DAE = \angle DBE$.

Since one side of $AEB$ of $\triangle DAE$ is produced, it follows from External Angle of Triangle is Greater than Internal Opposite that $\angle DEB$ is greater than $\angle DAE$.

But $\angle DAE = \angle DBE$, so $\angle DEB$ is greater than $\angle DBE$.

But $DB = DF$, so $DF$ is greater than $DE$.

But $DF$ is less than $DE$.

Thus we have a contradiction.

So the straight line $AB$ does not fall outside the circle.

Similarly, we can show that $AB$ does not lie on the circumference itself.

Hence the result.

$\blacksquare$