Draw a right-angled triangle with hypotenuse and . Let be on such that . Let and be on the perpendicular to at on either side of such that and .

Define the conchoid determined by and .

Let be the line from that forms an angle with the horizontal line . Let be the foot of the perpendicular from to .

Let be the intersection of and the conchoid. Let be the intersection of and . Then triangle is the solution.

By definition of the conchoid determined by the base with pole and the number , . By construction, . Since and are perpendicular projections of of the same size, is the angle bisector at and by construction .