Given the circle tangent to the sides $AB$ and $BC$, I want to construct another circle that is tangent to this circle and also tangent to the sides $AB$ and $AC$.
The center of such circle lies on the angle bisector of $\angle A$. Furthermore, the locus of the points that are equidistant from both $AB$ and the circle with center $O$ is a parabola with its focus at $O$ and the directrix parallel to $AB$ with the distance equal to the radius of the circle as mentioned here. Thus, the center of such circle can be found by intersecting the parabola and the angle bisector of $\angle A$ But I don't think it's possible to do this construction by using just a compass and a straight-edge.
Another thing that I've noticed is that when two circles are tangent to each other, their centers and their tangent point are placed on the same line. So, in order to do this construction it is enough to find the point of tangency and connect it to the center $O$ and extend it so that it intersects the angle bisector $A$ which gives the center of this circle. But I'm not sure how to find the point of tangency.
Best Answer
As noted in the comments and answers, this is a special case of the Problem of Apollonius, which dates back to antiquity. This special case is examined at cut-the-knot.org, from which this answer is adapted.
OP has already noted that the solution boils down to finding the point of tangency between the given and sought circles. What makes this possible is the observation that the point $S$ at which the circles touch would be a center of a homothety that takes either circle (and its tangents) to the other (and its tangents).
The construction is simple. We construct tangents to the given circle that are parallel to the sides $AC$ and $AB$. These tangents meet at $H$, and the line $AH$ consists of centers of homotheties that take tangents $AB$ and $AC$ of the sought circle to the constructed tangents of the given circle (and vice versa). There are two intersections of $AH$ with the given circle. One intersection is on $BC$ and leads to a trivial solution. But the other intersection $S$ gives us the point of contact between the two circles, and the intersection of $OS$ with the angle bisector of $\angle A$ gives us $K,$ the center of the sought circle.