Construct a triangle if a, ha and mb±mc are given.

This problem was proposed by Luís Lopes in Hyacinthos newsgroup

Suppose that the solution ABC has been found. Consider the triangle CDE obtained by parallel translation, i.e, DA is parallel to BC and A is the midpoint of BE. We call M, N, L the midpoints of sides CD, DE and AC, respectively.

Then we have DA=BC=a, DC=2 mc and DE=2 mb, and this lead us to a solution using the locus tool of Cabri-Géomètre:

Fix a segment DA of length a. Draw two r and s parallel lines to DA, one opposite to another, at distances ha and (1/2)ha from DA.

Consider any point M laying in the line s. Let E be the intersection point of r and MA. Now, join MD, DE and construct a point L in DE such that MD + ML = mc+mb. The problem is finished if E was the simmetry point of D with respecto to L, i.e. if this simmetry point was in line r. This is not true for all M, but we can use then locus tool to find the locus of the symmetry point of D respect to L when M varies in s. Then, we can find the intersection points from this locus and line r.

One of these points is named in the figure as E, and it is the starting point for the construction: from D, A, s and E, we get M, B, C easily.