The angle bisector at A of triangle ABC cuts BC at L. If C describes a circle whose center is A and B remains fixed, what is the locus of the points L?
I tried some sketches and some geogebra but I don't know how to catch and land my idea. Any hint?
Thank you !!
As $AB$ is fixed and $AC=R$ doesn't change too, we can use $\frac{BL}{LC}=\frac{BA}{AC}$ thus it will be circle, homothetical to the given.
$$\frac{BC}{BL}=\frac{LC}{BL}+1=\frac{AC}{BA}+1=\frac{R+AB}{AB},$$ $$\frac{BL}{BC}=\frac{AB}{R+AB},$$ $$BL=\frac{AB}{R+AB}BC,$$ so the center of homothety is $B$ and the coefficient is $\frac{AB}{R+AB}$.