Constructing a rhombus $ABCD$ with $A$ a given point, $B$ on a given circle, and $C$ on a given line, such that $\angle A=60^\circ$

Question

Given a point $A$, a circle $b$ and a line $c$, construct a rhombus $ABCD$ such that $B \in b$, $C \in c$ and $ \measuredangle A = 60^\circ $.

Hello! I saw this geometry problem in a textbook and I tried to solve it, but I've made no progress so far. The thing is, I don't even know how to attack such a problem.

Any help appreciated!

Answer 1

Let us observe a rhombus $ABCD$ with $\measuredangle A = 60^\circ.$ Its diagonal $AC$ has the length $|AC|=|AB|\sqrt 3,$ the angle at $A$ in $\triangle ABC$ is of measure $30^{\circ}.$

Therefore, $C$ is an image of $B$ in a transformation $$\sigma=\rho \circ \theta,$$ where the common center of the rotation $\rho$ and the homothety $\theta$ is $A,$ the angle of rotation is $30^{\circ}$ and a coefficient of homothety is $\sqrt 3.$
Then $C$ is an intersection point of $b$ and $\sigma(c).$

For further solutions, we can consider an oppositely oriented angle of rotation.

The enclosed picture shows one solution, a second point $C'$ is marked but the rhombus is not constructed.