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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.

(Luís Lopes. Euclid#1203)

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 A_{1} and A_{2}.

The triangles A_{1}BC and A_{2}BC 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}).

Angel Montesdeoca. 2020 Nov. 18