Inscribing Square in Circle

Theorem

In the words of Euclid:

In a given circle to inscribe a square.

Let $ABCD$ be the given circle.

Let two diameters be drawn at right angles to one another.

Join $AB, BC, CD, DA$.

Then $\Box ABCD$ is the required square.

We have that $BE = ED$, $EA$ is common and $\angle BEA = \angle DEA$ are right angles.

So from Triangle Side-Angle-Side Equality, $\triangle ABE = \triangle ADE$ and so $AB = AD$.

For the same reason $BC = CD = AD$ and so all four sides $AB, BC, CD, DA$ are equal.

So $\Box ABCD$ is equilateral.

Next we have that $BD$ is a diameter of circle $ABCD$.

For the same reason, $\angle ABC$, $\angle BCD$ and $\angle ADC$ are also all right angles.

So by definition, $\Box ABCD$ is a square.

$\blacksquare$