In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$ 
Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.
Is the triangle $KLM$ also an equilateral side?  In what relation are the area of ​​triangle $KLM$ the area of ​​triangle$ ABC$?



My attempt: 
I can see that $KLM$ is an equilateral triangle.  But, why the fraction $\frac{1}{7}$ is the area ratio of ​​the triangle $KLM$ to the ​​triangle $ABC$?

Here is a proof without using Menelaus's thm.  Let $A_2$, $B_2$, and $C_2$ be the midpoints of $CA_1$, $AB_1$, and $BC_1$, resp.  Draw $MA_1$, $KA_2$, $KB_1$, $LB_2$, $LC_1$, $LC_2$, $MA_2$, $KB_2$, and $LC_2$.  In this proof, $[\mathcal{P}]$ is the area of a polygon $\mathcal{P}$.

Due to symmetry $\triangle KLM$ is equilateral.  Therefore $\angle C_1LA=\angle KLM=60^\circ$.  Since $AC_1=AB_2$, $\angle C_1B_2A=60^\circ$.  Therefore $\square AC_1LB_2$ is cyclic so we see that $\angle B_2LA=\angle B_2C_1A=60^\circ$.  That is $\angle KLB_2=180^\circ-\angle KLM-\angle B_2LA=60^\circ$.
By symmetry $\angle CKB_1=\angle C_1LA=60^\circ$ too.  Hence $\angle CKB_1=\angle KLB_2$, so $KB_1\parallel LB_2$.  Since $CB_1=B_1B_2$, we get $LK=KC$.  
Turn $\triangle KB_1C$ around $C$ in the counterclockwise direction until $CB_1$ coincides with $CA_2$.  Suppose $K_c$ is the image of $K$ under this transformation.  Then $\triangle CKK_c$ is equilateral with side length $KC=KL$.  This shows that
$$[CB_1KA_2]=[CKK']=[KML].$$
Now $[KB_1B_2]=[KB_1C]$ because $\triangle KB_1B_2$ and $\triangle KB_1C$ share the same altitude from $K$ and have the same base legth $B_1B_2=B_1C$.  Furthermore $\triangle KLB_2\cong \triangle CKA_2$ because $KL=CK$, $\angle KLB_2=60^\circ = \angle CKA_2$, and $LB_2=KB_2$ (by symmetry).  Therefore
$$[KB_1B_2L]=[KB_1B_2]+[KLB_2]=[KB_1C]+[CKA_2]=[CB_1KA_2].$$
By symmetry, we have
$$[AC_1LB_2]=[BA_1MC_2]=[CB_1KA_2]$$
and
$$[MA_1A_2K]=[KB_1B_2L]=[LC_1C_2M].$$
Therefore $[KLM]=\frac{1}{7}[ABC]$.

Here is another argument why $\triangle KML$ is equilateral (which is not just saying "due to symmetry").  Note that $\triangle ABA_1$ is obtained by a rotation of $\triangle CAC_1$ about the center of $\triangle ABC$ by $60^\circ$ in the counterclockwise direction.  Therefore $CC_1$ and $AA_1$ make an angle of $60^\circ$.  Hence $\angle KLM=60^\circ$.  By the same argument $\angle LMK=60^\circ$ and $\angle MKL=60^\circ$ too.  

If you want to use a Menelaus argument, you can do it like this.  From Menelaus's thm (on $\triangle ABA_1$ with the traversal $C_1LC$)
$$\frac{AC_1}{C_1B}\cdot\frac{BC}{CA_1}\cdot \frac{A_1L}{LA}=-1.$$
Here the length ratios are signed (the ratio of lengths measured in the same direction is positive, and the ratio of lengths measured in the opposite directions is negative).  That is
$$\frac{1}{2}\cdot\left(-\frac{3}{2}\right)\cdot\frac{AL_1}{LA}=-1\implies \frac{AL_1}{LA}=\frac{4}{3}.$$
Using Menelaus's thm on $\triangle AA_1C$ with traversal $BMB_1$ yields
$$\frac{AM}{MA_1}\cdot \frac{A_1B}{BC}\cdot\frac{CB_1}{B_1A}=-1.$$
Hence
$$\frac{AM}{MA_1}\cdot\left(-\frac{1}{3}\right)\cdot\frac{1}{2}=-1\implies \frac{A_1M}{MA}=\frac{1}{6}.$$
This shows that $A_1M:ML:LA=1:3:3$.
We have
$$\frac{[KLM]}{[LA_1C]}=\frac{LM}{LA_1}\cdot \frac{LK}{LC}.$$
By symmetry $\frac{LK}{LC}=\frac{ML}{MA}=\frac12$, so
$$\frac{[KLM]}{[LA_1C]}=\frac{3}{3+1}\cdot\frac{1}{2}=\frac{3}{8}.$$
Similarly
$$\frac{[LA_1C]}{[C_1BC]}=\frac{CA_1}{CB}\cdot\frac{CL}{CC_1}=\frac{2}{3}\cdot\frac{3+3}{1+3+3}=\frac{4}{7}.$$
Finally
$$\frac{[C_1BC]}{[ABC]}=\frac{BC_1}{BA}=\frac{2}{3}.$$
Therefore
$$\frac{[KLM]}{[ABC]}=\frac{[KLM]}{[LA_1C]}\cdot \frac{[LA_1C]}{[C_1BC]}\cdot \frac{[C_1BC]}{[ABC]}=\frac{3}{8}\cdot\frac{4}{7}\cdot\frac{2}{3}=\frac17.$$
