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 $\triangle ABC$
$\endgroup$2 Answers
$\begingroup$This is a special case of Routh's Theorem.
$\endgroup$ 2 $\begingroup$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 $[AJKF]=5/21[ABC]$.
And the same analysis give the same results for the two remaining areas.
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