Intersecting Circles III

Given a square and an arbitrary length $l$, we’re going to construct a rectangle with one of its edges being of length $l$ such that the areas of the square and the rectangle coincide.

In the following sketch, two circles with same radius $\overline{AC}$ intersect in such a manner that their respective centers are located on the other circle’s boundary.

$\mbox{Symmetrically intersecting circles}$

Then the hatched triangle $\bigtriangleup_{BPA}$ is isosceles and thus $$ \overline{AB} = 2 \overline{AC} \cos \alpha~. $$ For the same reason $$ \overline{AC} = \overline{PA} = 2 \overline{AD} \cos \alpha $$ and so $$ \overline{AB} \cdot \overline{AD} = \overline{AC}^{\, 2}~. $$ If we now say that $\overline{AC}^{\, 2}$ is the square and that $l = \overline{AD}$, then $\overline{AB}$ does the job. This is called finding the third proportion. It solves Euclidea’s level 8.2.