Harmonic quadrilateral question

Question

Let $ABCD$ be a quadrilateral with $\angle A+\angle C=60$ and $|AB|.|CD|=|BC|.|AD|$.Show that $|AB|.CD|=|AC|.|BD|$

Answer 1

Take a point $E$ so that triangle $DCE$ is equilateral and is on the same half-plane with $ABCD$ with respect to the line $CD$. Now $AB\cdot CD = BC\cdot DA$ together with $CD = CE$ implies $$\frac{AB}{BC}=\frac{DA}{DC} = \frac{DA}{CE}.$$ This leads to $$\frac{AB}{AD}=\frac{CB}{CE}.$$ Now $\angle BCE =60^{\circ} - \angle BCD = \angle BAD$. Hence triangles $BAD$ and $BCE$ are similar. Thus $$\frac{AB}{CB}=\frac{AD}{CE}=\frac{BD}{BE}.$$ Moreover, $\angle ABD = \angle CBE$ from the similarity of the two latter triangles. Therefore $\angle ABC = \angle DBE$. Combined with $\frac{AB}{CB}=\frac{BD}{BE}$, it implies that triangles $ABC$ and $DBE$ are similar. It follows from here that $$\frac{AB}{AC} = \frac{BD}{DE} = \frac{BD}{CD},$$ because $DE = CD$ since triangle $DCE$ is equilateral. Now cross-multiply the latter identity and obtain $AB \cdot CD = AC \cdot BD$.