Construction of the [i]proportionatrix secunda[/i] in a semicircle of diameter AG. Steps to find a point K of the curve from a point F of the semicircle: [list=1] [*]Semicircle of diameter AG and center D(diagram). [*]Semicircle of diameter DG. [*]A point F is selected on the large semicircle, FG is drawn cutting the small semicircle in H. [*]Perpendicular HI from H to AG. [*]Semicircle of center I and radius IG. [*]K is the intersection of this semicircle with FG. [*]L is the intersection of this semicircle with AG. FL is drawn. [/list] The bottom bar of the canvas allows to follow these steps. Once the last step of the construction is arrived at, click on the 'Trace' button: the trajectory of the red point will trace the curve when F moves on the semicircle. To stop the process, click on 'Stop Trace'.You can also draw the curve by selecting the point on the larger semicircle and moving it along the semicircle. To clean the canvas and get the initial configuration, click on the button with two circular arrows in the right upper corner of the canvas.
Once the curve is drawn, it is easy to prove that, for each K, AG/GF = GF/GL = GL/GK: it is enough to show that AGF, FGL and GKL are similar triangles. This statement is proved in Villalpando's proposition XI ([url]http://www.e-rara.ch/zut/content/pageview/3800973[/url]). Villalpando's [i]proportionatrix secunda[/i] can be used to find two mean proportionals between lines [i]a,b[/i], with [i]a>b[/i]. If we set AG = [i]a[/i] and the [i]proportionatrix secunda[/i] is met by a circle of center G and radius [i]b[/i], we obtain K, being GK = [i]b[/i]. F is obtained prolongating GK and cuttig it with the biggest semicircle; L is obtained tracing a perpendicular line from F to AG. An Archytas' solution is the figure ALGKF (Villalpando's actual use of this curve to find the two mean proportionals between two assigned straight lines is a bit different, still is equivalent to the one just expounded).