Let *ABCD *be the given circle, K the triangle
described

Then, if the circle is not equal to *K, it *must be either
greater or less.

I. If possible, let the circle be greater thanK.

Inscribe a square *ABCD, *bisect the arcs *AB, BC, CD,
DA, *then bisect (if necessary) the halves, and so on, until the sides
of the inscribed polygon whose angular points are the points of division
subtend segments whose sum is less than the excess of the area of the circle
over K.

Thus the area of the polygon is greater than K.

Let *AE* be any side of it, and *ON *the perpendicular
on *AE *from the centre *0.*

Then* ON* is less than the radius of the circle
and therefore less than one of the sides about the right angle in K. Also
the perimeter of the polygon is less than the circumference of the circle,
i.e. less than the other side about the right angle in
*K.*

Therefore the area of the polygon is less than *K*; which
is inconsistent with the hypothesis

Thus the area of the circle is not greater than
*K.*

II. If possible,_{ }let the circle be less than
*K*

Circumscribe a square,_{ }and let two adjacent sides,
touching the circle in* E*, *H, *meet in *T*. Bisect
the arcs between adjacent points of contact and draw the tangents at the
points of bisection. Let *A* be the middle point of the arc *EH*,
and *FAG *the tangent at *A.*

Then the angle *TAG *is a right angle.

Therefore

TG |
> | GA |

> | GH. |

It follows that the triangle *FTG* is greater than half
the area *TEAH.*

Similarly, if the arc *AH *be bisected and the tangent
at the point of bisection be drawn, it will cut off from the area *GAH*
more than one-half.

Thus, by continuing the process, we shall ultimately arrive at a circumscribed polygon such that the spaces intercepted between it and the circle are together less than the excess of K over the area of the circle.

Thus the area of the polygon will be less than
*K*

Now, since the perpendicular from *0* on any side of the
polygon is equal to the radius of the circle, while the perimeter of the
polygon is greater than the circumference of the circle, it follows that
the area of the polygon is greater than the triangle *K; *which is
impossible.

Therefore the area of the circle is not less than K.

Since then the area of the circle is neither greater nor less
than K, it is equal to it.

*The area of a circle is to the square on its diameter as
11 to 14*

[The text of this proposition is not satisfactory, and Archimedes
cannot have placed it before Proposition 3, as the approximation depends
upon the result of that proposition.]

*The ratio of the circumference of any circle to its diameter
is less than 3 1/7
but greater than 3
10/71.*