Areas of Circles are as Squares on Diameters

Proof

 * Euclid-XII-2.png

Let $ABCD$ and $EFGH$ be circles.

Let $BD$ be a diameter of $ABCD$.

Let $FH$ be a diameter of $EFGH$.

It is to be demonstrated that the ratio of the area of $ABCD$ to the area of $EFGH$ equals the ratio of the square on $BD$ to the square on $FH$.

Suppose to the contrary that the square on $BD$ to the square on $FH$ does not equal the area of $ABCD$ to the area of $EFGH$.

Then:
 * $BD^2 : FH^2 = ABCD : S$

where $S \ne EFGH$.

Suppose WLOG that $S < EFGH$.

Let the square $EFGH$ be inscribed in the circle $EFGH$.

The inscribed square is half the area of a square which has been circumscribed about the circle $EFGH$.

Therefore the circle $EFGH$ is greater than the square $EFGH$ but less than twice the square $EFGH$.

Hence square $EFGH$ is greater than half the circle $EFGH$.

Let the arcs $EF, FG, GH, HE$ be bisected at $K, L, M, N$.

Let $EK, KF, FL, LG, GM, MH, NE$ be joined.

Let the parallelogram be completed on $\triangle EKF$.

Then $\triangle EKF$ is half of this parallelogram.

Therefore the segment $EKF$ of the circle $EFGH$ is less than twice $\triangle EKF$.

Hence each of $\triangle EKF, \triangle FLG, \triangle GMH, \triangle HNE$ is greater than half the segment of the circle $EFGH$ around it.

This process can be repeated.

We continue to bisecting the remaining arcs and joining straight lines repeatedly.

By :
 * eventually we will leave some segments of the circle $EFGH$ which will be less than the amount by which the circle $EFGH$ exceeds $S$.

Therefore the remainder, the polygon $EKFLGMHN$ is greater than the area $S$.

Let the polygon $AOBPCQDR$ similar to $EKFLGMHN$ be inscribed in the circle $ABCD$.

Therefore from :
 * $BD^2 : FH^2 = AOBPCQDR : EKFLGMHN$

But:
 * $BD^2 : FH^2 = ABCD : S$

So by :
 * $ABCD : S = AOBPCQDR : EKFLGMHN$

So by :
 * $ABCD : AOBPCQDR = S : EKFLGMHN$

But:
 * $ABCD > AOBPCQDR$

Therefore:
 * $S > EKFLGMHN$

But we started with the premise that:
 * $S < EKFLGMHN$

Therefore the square on $BD$ to the square on $FH$ does not equal the area of $ABCD$ to any area less than $EFGH$.

Using the same technique we can also show that the square on $FH$ to the square on $BD$ does not equal the area of $EFGH$ to any area less than $ABCD$.

It is now to be shown that the ratio of the area of $ABCD$ to an area greater than $EFGH$ can equal the ratio of the square on $BD$ to the square on $FH$.

Suppose to the contrary, that:
 * $ABCD : S = BD^2 : FH^2$

where $S > EFGH$.

that is:
 * $FH^2 : DB^2 = S : ABCD$

But as $S$ is to the circle $ABCD$, so $EFGH$ is to an area less than the circle $ABCD$.

Therefore by :
 * the ratio of the square on $BD$ to the square on $FH$ equals the ratio of the circle $EFGH$ to an area less than the circle $ABCD$.

This was shown to be impossible.

Therefore the square on $BD$ to the square on $FH$ does not equal the area of $ABCD$ to any area greater than $EFGH$.

Hence the result.