This Demonstration shows Vieta's solution of the depressed cubic equation , where . To solve it, draw an isosceles triangle with base and unit legs. Let be the angle at the base and . Draw a second isosceles triangle with base angle and unit legs. The base of the second triangle is a root of the equation.

Proof

By construction, , . Let . Since and are similar, . The perpendicular projections of and on also give similar right-angled triangles. So . So and . Eliminating gives .

Since , and applying the law of cosines to , we find .

[1] G. E. Martin, Geometric Constructions, New York: Springer, 1998 pp. 126–140.

[2] M. Hladnik, "Some Historical Constructions of the Regular Heptagon" (in Slovenian), Obzornik za matematiko in fiziko, 61(4), 2014 pp. 132–145. www.obzornik.si/61.