:800px 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 $EFGH$. From Square Inscribed in Circle is greater than Half Area of Circle: :the square $EFGH$ is greater than half the circle $EFGH$. Let the arcs $EF, FG, GH, HE$ be bisected at $K, L, M, N$. Let $EK, KF, FL, LG, GM, MH, HN, NE$ 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 {{EuclidPropLink|book = X|prop = 1|title = Existence of Fraction of Number Smaller than Given}}: : 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 {{EuclidPropLink|book = XII|prop = 1|title = Areas of Similar Polygons Inscribed in Circles are as Squares on Diameters}}: :$BD^2 : FH^2 = AOBPCQDR : EKFLGMHN$ But: :$BD^2 : FH^2 = ABCD : S$ So by {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$ABCD : S = AOBPCQDR : EKFLGMHN$ So by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$ABCD : AOBPCQDR = S : EKFLGMHN$ 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 {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :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. {{qed}} {{Euclid Note|2|XII}}