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

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

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

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

• 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 (𝒫_b) 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 (𝒫_b).

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

Tangents at C to (𝒫_b) 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 (𝒫_c). The tangents at B to (𝒫_c) coincide with those obtained in the case of the parabola (𝒫_b).