Rectangle Squaring

Given a Rectangle $\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern10.6pt \vrule width.6pt} \hrule height.6pt}BCDE$, draw $EF=DE$ on an extension of $BE$. Bisect $BF$ and call the Midpoint $G$. Now draw a Semicircle centered at $G$, and construct the extension of $ED$ which passes through the Semicircle at $H$. Then $\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern6.4pt \vrule width.6pt} \hrule height.6pt}EKLH$ has the same Area as $\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern10.6pt \vrule width.6pt} \hrule height.6pt}BCDE$. This can be shown as follows:

$$A(\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern10.6pt \vrule width.6pt} \hrule height.6pt}BCDE) = BE\cdot ED = BE\cdot EF$$

$$= (a+b)(a-b)=a^2-b^2=c^2.$$

References

Dunham, W. ``Hippocrates' Quadrature of the Lune.'' Ch. 1 in Journey Through Genius: The Great Theorems of Mathematics. New York: Wiley, pp. 13-14, 1990.