I think that if you say that "AB is on the ray e, the point C is on the ray f" you must say also that the angle is acute, btw i find a solution using the rectangolar hyperbola.

Interesting hyperbola...

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this was my solution: lemma : Let a triangle ABC, then the locus of the isogonal conjugate of the perpendicolar axis of BC is a rectangolar hyperbola with center the midpoint of BC and asimptote the internal and external angle bisector of the medial triangle from the midpoint of BC. let H and K the projection of P and Q on E. So now apply the lemma to the degenetete triangle formed by the sideline PQ, PH and QK we get that the locus of the vertex of all isosceles triangle with base on e and P on a side and Q on the other is a rectangolar hyperbola with center the midpoint of PQ and asimptote perpendicolar and parallel to e. Then C stay of the intersection of this hyperbola with f. This is a costruction with ruler and a compass of the intersection of a conic C with a line L: Let F be a focus, l a directrix, and e = the eccentricity. (1) Let $ H = L \cap l$ (2) Take an arbitrary point P with pedal Q on the directrix. (3) Construct a circle, center P, radius e · PQ. (4) Through P construct the parallel to L, intersecting the directrix at O. (5) Through O construct the parallel to FH, intersecting the circle above in X and Y . (6) The parallels through F to PX and PY intersect the given line L at two points on the conic.

Construct Q', the reflection of Q in ray f. Then If $\alpha$ is the angle between e and f, we have that $\angle PCQ' = \angle PCO + \angle OCQ' = \angle PCO +\angle OCQ = \angle QCP + 2 \angle PCO = \angle BCA + 2 \angle ACO= \angle BCA + 2 (\angle BAC - \alpha) = 180 - 2 \alpha.$ So to find C, draw the circular arc PQ' (away from O) with angle $180 - 2 \alpha$, and C is the intersection of that arc with f.