Quadrilateral $ABCD$ with $AB=AD$, $\angle BAD=60^\circ$, $\angle BCD=120^\circ$. Prove $BC+DC=AC$

Question

In a given quadrilateral $ABCD$, we have $$AB = AD, \angle BAD = 60^\circ, \angle BCD = 120^\circ$$ Prove that $$ BC + DC = AC$$

I know the quadrilateral is cyclic. I have been able to solve this for the special case where $C$ is the midpoint of arc $BD$, but I am not sure how to generalize.

Answer 1

Hint: $ABD$ is an equilateral triangle.

Hint: $ABCD$ is a cyclic quad.

The result follows by applying Ptolemy's theorem.

Answer 2

Continue with the cyclic ABCD, as well as the equilateral triangle ABD.

Locate the point E on AC such that CD = CE. Since $\angle$ACD = $\angle$ABD = 60, CDE is equilateral and CD = DE. Along with AD = BD and $\angle$CAD = $\angle$CBD, we have congruent triangles ADE and BDC. Thus, AE = BC and

$$AC = AE + EC = BC + CD$$.