Let B be an arbitrary point on circle a. Perform homothety (central similarity transformation) with center B and coefficient of similarity a/b on circle b. We'll get the blue circle with radius a. The blue circle intersects circle lc at point Lc. Draw the line BLc and continue it beyond Lc to the intersection with the circle b at point A. ABC solves our problem.

Indeed, its two sides AC and BC have the given lengths b and a, respectively. Segment CLc has the length lc, and the only thing we have to show is that CLc is the bisector of angle C. We know that, for the bisector, the point Lc divides the side AB in the ratio a/b. Since, on AB, there is only one such point, the point defines the bisector uniquely. But, by the costruction, our point Lc has exactly that property. (Points that are mapped into each other by homothety lie on a straight line through the center of homothety, B in our case. Their distances to B are in the given ratio.) Therefore, by construction, the line CLc is indeed the bisector of angle C.

Now I will show my ( supposed ) solution:

a, b, lc
With the center C draw three circles of radii a, b, and lc. For the simplicity sake, the circles will be denoted by their radii: circle a, circle b, and circle lc. We are looking for a segment AB (with A on the circle b and B on the circle a) that is cut by the circle lc in the ratio a/b (recollect one of the basic properties of angular bisectors.)

Let B be an arbitrary point on circle a. Perform homothety (central similarity transformation) with center B and coefficient of similarity a/(a+b) on circle b. We'll get the blue circle with radius a’. The blue circle intersects circle lc at point Lc. Draw the line BLc and continue it beyond Lc to the intersection with the circle b at point A. ABC solves our problem.

and ( point C is not a point of the blue circle, only in the case if a = lc )

BLc = AB(a/a+b) and LcA = AB(b/a+b) , so BLc /LcA = a/b

Indeed, its two sides AC and BC have the given lengths b and a, respectively. Segment CLc has the length lc, and the only thing we have to show is that CLc is the bisector of angle C. We know that, for the bisector, the point Lc divides the side AB in the ratio a/b. Since, on AB, there is only one such point, the point defines the bisector uniquely. But, by the costruction, our point Lc has exactly that property. (Points that are mapped into each other by homothety lie on a straight line through the center of homothety, B in our case. Their distances to B are in the given ratio.) Therefore, by construction, the line CLc is indeed the bisector of angle C.

Please, will you send me your answer with your opinion.

Thanks !

J. Vriens