*Originally posted by Acolyte*

**That's impressive! How many steps does each construction take?**

The iterative method which I have found , is as follows:

Let ACB be the given circle with centre O, AB its (horizontal) diameter, and C is a point on its circumference such that OC is perpendicular on diameter AOB.

With ruler and compass, draw /_BOP1 = 22.5 deg. Take point P1 on OP1=BC in length. Draw perpendicular P1Q1 at P1 on OP1 which meets extended line OB at Q1. Now draw bisector OP2 of the /_BOP1 & take point P2 on this bisector such that OP2=OQ1.

Now draw perpendicular P2Q2 at P2 on the line OP2 which meets the extended line OB at point Q2. Next draw bisector OP3 of the /_BOP2. And go on doing this iterative construction and locating the successive points Q1, Q2. Q3....on the extended line OB.

In fact the points Q1, Q2, Q3,..converge so fast that Q3 is practically indistinguishable from Q2. The final point Q of convergence is practically reached after the second iteration .

Next extend OA backwards upto A1 so that AA1=OA, i.e. A1O=2OA. Now draw a semi-circle A1XQ with A1Q as the diameter.

Now let the extended line OC meet this cicle at point X.

Now draw a square on OX. This square is the required square equl to the given circle in area.All this can be done with a ruler & compass .

This iterative construction uses the trigonometrical identity

Z = sin Z*{sec Z/2 * sec Z/4 * sec Z/8......sec Z/(2^n).....to infinity}.