Chord Lies Inside its Circle

Theorem
If on the circumference of a circle two points be taken at random, the straight line joining the points (i.e. the chord joining these points) will fall within the circle.

Geometric Proof

 * Euclid-III-2.png

Let $$ABC$$ be a circle where $$A, B$$ are two points taken at random on its circumference.

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).

Find the center $$D$$ of circle $$ABC$$.

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

As $$DA = DB$$ then from Isosceles Triangles have 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 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 from Greater Angle of Triangle Subtended by Greater Side, $$DB$$ is greater than $$DE$$.

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.