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.
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 .