Find the area of three quadrilaterals inside an equilateral triangle $\triangle ABC$ is an equilateral triangle with side length of $1$. $D,E,F$ are $\frac{1}{3}$ away from $C,A,B$. What is the total area of the three quadrilaterals 1,2 and 3 enclosed by the orange sides? I can think about finding the coordinates of the vertices of the quadrilateral and use formula to calculate the area (see note below), but could not find an easy geometric way to solve this.

Note:
Solution by coordinates calculation
AF intercepts EC at G, BD at H, and BD intercept EF at I.
Given  $A(0, \frac{\sqrt 3}{2}, B(-\frac{1}{2},0), C(\frac{1}{2},0)$,
$E(x_1,y_1)=E(-\frac 16,\frac{\sqrt3}{3}),G(x_2,y_2)=G(-\frac 1{14},\frac{2\sqrt3}{7}),H(x_3,y_3)=H(-\frac 17,\frac{\sqrt3}{14}),I(x_4,y_4)=I(-\frac 16,\frac{\sqrt3}{15})$
$S_1=S_2=S_3=\frac {1}{2} ⋅ {(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1)
- (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)}\approx 0.02612$
$S=S_1+S_2+S3 \approx 0.07835$
$\frac{S}{\triangle ABC} \approx 0.18$
 A: 
$Area\left(EGHK\right)=Area\left(EJD\right)-2Area\left(EJG\right)$
Applying Menelaus Theorem on $\triangle CEB$ with $DGA$ as the transversal will give $\frac {CG}{GE}$. Applying Menelaus Theorem on $\triangle CEF$ with $GJA$ as the transversal will give $\frac {EJ}{JF}$.
Now, $\frac {Area\left(EJD\right)}{Area\left(FJD\right)}=\frac {EJ}{JF}$ and hence $Area\left(EJD\right)$ is known.
Applying Menelaus Theorem on $\triangle FJD$ with $LGE$ as the transversal gives the value of $\frac {DG}{GJ}$ and thereafter like we did before, $Area\left(EJG\right)$ is known.
$Area\left(DEF\right)$ can be derived by subtracting the areas of the three congruent triangles from the original $\triangle ABC$.
Hence, plugging these areas into the equation mentioned at the very beginning will give $Area\left(EGHK\right)$
A: 
$$A_{AEC}=\frac 13 A_{ABC}$$
$$A_{CED}=\frac 13 A_{AEC}$$
$\Rightarrow A_{CED}=\frac 19 A_{ABC}$
$A_{DHGI}\approx \frac 12 A_{CED}=\frac 1{18}A_{ABC}$
Therefore the sum of three quadrilaterals 1, 2 and 3 is about $\frac 16 A_{ABC}$.
A: Denote area of $\triangle ABC$ by $\Delta$. Let areas of the three quadrilaterals each be $X$ and the blue triangles in following diagram, each be $Y$.

We'll take ratios approach, which is fun, form two linear equations in $X,Y$ and solve for it.
First notice $\triangle$s $AED$, $BFE$, $CDF$ are all $30^\circ-60^\circ-90^\circ$. Each of these e.g., $\triangle BFE$ has base one-third and height two-third the original, hence area of equilateral $\triangle DFE$ is
$$[DFE] = \Delta \left(1-3\cdot \frac{1}{3} \cdot \frac{2}{3}\right)=\frac{\Delta}{3}$$
If you don't know the hatched triangle is the one-seventh area triangle, we can also activate Routh's theorem to obtain its area as $\Delta/7$. We write
$$[DFE] = \frac{\Delta}{3} = 3X + 3Y + \frac{\Delta}{7} \tag{1}$$
Next $\triangle AFE$ has same height but one-third base of $\triangle ABF$ which in turn has one-third base of $\triangle ABC$, so $[AFE] = \Delta/9$.
Drop $FP \perp AB$. $\triangle BPF$ is also $30^\circ-60^\circ-90^\circ$. Let $BF=a$. $\triangle AQE \sim \triangle AFP$ with ratio $AE/AP=AE/(AB-BP)=2/5$. Thus $AQE$ has same base but two-fifths the height of $\triangle AFE$. We can write
$$[AFE]=\frac{\Delta}{9}=X+2Y+\frac{2}{5}\cdot \frac{\Delta}{9} \tag{2}$$
On simplifying $(1),(2)$,
$$3X+3Y=\frac{4}{21}\Delta \tag{3}$$
$$X+2Y=\frac{1}{15}\Delta \tag{4}$$
Eliminating $Y$, we get
$$3X=\left(2\cdot \frac{4}{21} - 3\cdot  \frac{1}{15} \right) \Delta$$
That is
$$\boxed{\text{Area of quadrilaterals} = \frac{19}{105}\Delta}$$
which is about $18 \%$ of area of $\triangle ABC$.
