How to find area of a quadrilateral in complex plane?

Question

Let $ABCD$ be a convex quadrilateral where $AB = AD$. Also given that $BC + DC = AC = 1$ and $\angle A = 60^{\circ}$. Find the area of the polygon. I've been told that this problem has something to do with a complex plane. I let the center of the plane to be in $A$ and real axis along bisector of $\angle BAD$, imaginary axis perpendicular to it. Now if a point $B$ has a complex number $z$ then $D$ is $\overline{z}$ because $\angle A$ is $60^{\circ}$ and $AB = AD$ hence it's equilateral triangle and points $B$ and $D$ are symmetric across the real axis. If point $C$ has a complex number $x$, then $x\overline{x} = 1$ because it lies on the unit circle and $|x - z| + |x - \overline{z}| = 1$. However from here I struggle to derive $x$ through $z, \overline{z}$ and find the area of $ABCD$. Any suggestions on what to do next?