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 $\triangle ABC$


This is a special case of Routh's Theorem.

  • $\begingroup$ That solves the last problem, but what about the rest? $\endgroup$ – Stefan4024 Feb 14 '14 at 3:14
  • $\begingroup$ @Stefan4024 Take a good look to the proof, the key to solve the rest is there. $\endgroup$ – chubakueno Feb 14 '14 at 5:06

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