Construct a triangle given a, ha and such that the A-internal bisector (or the A-external bisector) have the direction of a vector v.

(Luís Lopes. Euclid#1203)

On a line, d, parallel to side BC (given) at a distance ha, we take an arbitrary point M.

The reflection of the line MB in the line through M parallel to v, involves a parabola (𝒫b).

• The line p through B parallel to v is tangent to (𝒫b).

• Let D be d ∩ p and let T be the intersection of the perpendicular bisector of BD and the line d. The line d is tangent to (𝒫b) at T.

• The line through C and parallel to v intersects d at a point E. The line q, reflections of EB in EC, is tangent to (𝒫b).

With these three data the parabola (𝒫b) can be constructed (see section 13.2 (p.346) "A Gallery of Conics by Five Elements". Paris Pamfilos).

Tangents at C to (𝒫b) intersect d at points A1 and A2.

The triangles A1BC and A2BC are the solutions to proposed problem. In one of them v is parallel to the A-interior bisector, and in the other v is parallel to the A-exterior bisector.

If instead of taking the line MB, we take the line MC and if we proceed analogously, we obtain another parabola (𝒫c). The tangents at B to (𝒫c) coincide with those obtained in the case of the parabola (𝒫b).

Angel Montesdeoca. 2020 Nov. 18