Proof: By Euclid
(related to Proposition: 4.06: Inscribing Square in Circle)
- Let two diameters of circle $ABCD$, $AC$ and $BD$, have been drawn at right angles to one another. And let $AB$, $BC$, $CD$, and $DA$ have been joined.
- And since $BE$ is equal to $ED$, for $E$ (is) the center (of the circle), and $EA$ is common and at right angles, the base $AB$ is thus equal to the base $AD$ [Prop. 1.4].
- So, for the same (reasons), each of $BC$ and $CD$ is equal to each of $AB$ and $AD$.
- Thus, the quadrilateral $ABCD$ is equilateral.
- So I say that (it is) also right-angled.
- For since the straight line $BD$ is a diameter of circle $ABCD$, $BAD$ is thus a semicircle.
- Thus, angle $BAD$ (is) a right angle [Prop. 3.31].
- So, for the same (reasons), (angles) $ABC$, $BCD$, and $CDA$ are also each right angles.
- Thus, the quadrilateral $ABCD$ is right-angled.
- And it was also shown (to be) equilateral.
- Thus, it is a square [Def. 1.22] .
- And it has been inscribed in circle $ABCD$.
- Thus, the square $ABCD$ has been inscribed in the given circle.
- (Which is) the very thing it was required to do.
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"