Extended Golden Triangle

$\triangle ABC$ can be stretched and skewed to an equilateral triangle with a side length of $1$ and still preserve the ratio of areas:

Using the law of cosines on $\triangle ADF$ , $DF = \sqrt{\phi^2 + (\phi -1)^2 - 2 \cdot \phi \cdot (\phi - 1) \cdot \cos 120°} = 2$ , and by symmetry $\triangle DEF$ is also an equilateral triangle.

Since the ratio of sides of the two equilateral triangles is $\frac{2}{1}$ , the ratio of their areas is $\frac{2^2}{1^2} = \boxed{4}$ .

Labelling $a=BC$ , $b=CA$ , $c=AB$ , we have $AD=\varphi c$ and $AF=(\varphi -1)b$ . Hence $\text{Area }\Delta AFD = \frac12 \varphi (\varphi-1) bc \sin A= \varphi (\varphi-1) \times \text{Area }\Delta ABC$

Since by definition, $\varphi (\varphi-1)=1$ , the areas of the four triangles $\Delta ABC,\Delta AFD,\Delta BDE,\Delta CEF$ are all the same; so the area of $\Delta DEF$ is $\boxed4$ times the area of $\Delta ABC$ .

Note that $\overrightarrow{BE} = \varphi\overrightarrow{BC}$ and $\overrightarrow{BD} = (\varphi-1)\overrightarrow{AB}$ , so that $\overrightarrow{BD} \times \overrightarrow{BE} \; = \; \varphi(\varphi-1)\overrightarrow{BC} \times \overrightarrow{BA} \; = \; \overrightarrow{BC} \times \overrightarrow{BA}$ and hence triangles $ABC$ and $BDE$ have the same area. SImilarly, $AFD$ and $CEF$ have the same area as well, and hence the ratio between the areas of $DEF$ and $ABC$ is $\boxed{4}$ .