Area of a derived triangle

We have $\Delta ABC = \frac12 CA \cdot BC \sin C$ and $\Delta CEF=\frac12 CE \cdot CF \sin C=\frac19 CA \cdot BC \sin C$

so (by symmetry) $\Delta CEF=\Delta BDE=\Delta AFD=\frac29\Delta ABC$

and $\Delta DEF=\Delta ABC - 3\times \frac29\Delta ABC = \frac13 \Delta ABC$

so the answer is $\boxed{\frac13}$ .

Since the required ratio is independent of the shape of the triangle $ABC$ , let us take this triangle as a right isosceles triangle, right angled at $A$ . Let the coordinates of $A, B$ and $C$ be $(0,0),(3a, 0)$ and $(0,3a)$ respectively.

Then those of $D, E$ and $F$ are $(a, 0),(2a,a)$ and $(0,2a)$ respectively.

So, area of $\triangle {DEF}$ is $\dfrac {3a^2}{2}$ ,

and that of $\triangle {ABC}$ is $\dfrac {9a^2}{2}$

Therefore the required ratio is $1:3\approx \boxed {0.3333}$ .

Let the side lengths opposite $A$ , $B$ , and $C$ be $a$ , $b$ , and $c$ ; and the respective altitudes be $h_a$ , $h_b$ , and $h_c$ . Then the area of $\triangle ABC$ , $[ABC] = \dfrac {ah_a}2 = \dfrac {bh_b}2 = \dfrac {ch_c}2$ .

We note that $[CEF] = \dfrac 12 \cdot \dfrac 23 a \cdot \dfrac 13 h_a = \dfrac {ah_a}9 = \dfrac 29 [ABC]$ . Similarly $[ADF] = \dfrac {bh_b}9 = \dfrac 29 [ABC]$ and $[BDE] = \dfrac {ch_c}9 = \dfrac 29 [ABC]$ .

Therefore,

$\frac {[DEF]}{[ABC]} = \frac {[ABC]-[CEF]-[ADF]-[BDE]}{[ABC]} = 1 - \frac 23 = \frac 13 \approx \boxed{0.333}$

$\triangle ABC$ can be stretched and skewed to an equilateral triangle with a side length of $3$ and still preserve the ratio of areas:

Using the law of cosines on $\triangle ADF$ , $DF = \sqrt{1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 60°} = \sqrt{3}$ , and by symmetry $\triangle DEF$ is also an equilateral triangle.

Since the ratio of sides of the two equilateral triangles is $\frac{\sqrt{3}}{3}$ , the ratio of their areas is $\frac{\sqrt{3}^2}{3^2} = \frac{1}{3} \approx \boxed{0.3333}$ .