Golden Ratio in Two Equilateral Triangles

Here a most elegant construction of the Golden Ratio posted by Tran Quang Hung (with a proof) at the CutTheKnotMath facebook page.

Construction

$ABCD$ is a rhombus with $2AC=BD;$ $(O)$ the inscribed circle; $E$ and $F$ the points of intersection of $(O)$ with $BD.$

Then $F$ divides $DE$ in the golden ratio.

Proof

Let $M$ be the point of tangency of $(O)$ with $AD.$

Triangles $OMD$ and $AOD$ are similar. It follows that $\displaystyle\frac{MD}{MO}=\frac{OD}{OA}=2$. Thus, $MD=2OM=EF.$ By a property of tangent, $DM^2=DF\cdot DE.$ From this, $EF^2=DF\cdot DE,$ or $\displaystyle\frac{FD}{FE}=\frac{EF}{ED},$ as required.