# Area of triangle inside triangle

In triangle $$ABC$$ we choose 3 points $$D,E,F$$, such that $$\overline{AD} = \frac 13 \overline{AB}, \overline{BE} = \frac 13 \overline{BC}, \overline{CF} = \frac 13 \overline{CA}$$. Draw segments $$\overline{CD}, \overline{BF}, \overline{CD}$$, like in the picture.

Now prove that $$A_{DJA} = A_{BLE} = A_{CKF} = \frac 13 A_{KJL}$$. Prove that quadrilaterials $$AJKF$$, $$DJLB$$ and $$KLEC$$ have same area. And at last prove that $$A_{KLJ} = \frac 17 A_{ABC}$$.

I've only managed to prove that the sum of the areas of the smaller triangles is the same as the area of $$\triangle KLJ$$, but nothing more. I've tried to use the fact that $$A_{ABE} = A_{ACD} = A_{BFC} = \frac 13 A_{ABC}$$, but it didn't help me.

P.S. $$A_{ABC}$$ represents the area of $$\triangle ABC$$

In the diagram, we wish to first compute the area ratios of the three triangles surrounding the inner triangle $$\triangle JKL$$ using Menelaus' Theorem. First consider $$\triangle ABE$$ and the Menelaus line segment $$DC$$ passing through it. We then have $$\frac{BC}{CE}\times \frac{JE}{JA}\times \frac{AD}{DB}=1$$ or $$\frac{JE}{JA}=4/3$$ based on the cevian ratios given. This means, using area ratios, that $$[JEC]=4/3 [AJC]$$. Letting $$x=[AJC]$$ then $$7/3x=[AEC]$$ or $$[AJC]=3/7[AEC]$$. Since $$\frac{EC}{EB}=2$$, we now let $$y=[ABE]$$ then $$2y=[AEC]$$ or $$3y=[ABC]$$ so that $$[ABE]=1/3[ABC]$$ so $$[AEC]=2/3[ABC]$$. Finally, we obtain $$[JAC]=(3/7)(2/3)[ABC]=6/21[ABC]$$.
And we can do an identical analysis for the remaining triangles $$\triangle BKC$$ and $$\triangle ABL$$ and obtain the same $$6/21[ABC]$$ for each area giving $$[JKL]=1/7[ABC]$$.
In order to compute the areas of the blue triangles, we again use similar ratios, we have $$1/3[ABC]=[BKC]+[KFC]$$ or from the results above, $$1/3[ABC]=3/21[ABC]+[KFC]$$ so that $$[KFC]=1/21[ABC]$$ leaving the quadrilateral $$[AJKF]=5/21[ABC]$$.