2
$\begingroup$

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 $\triangle ABC$

$\endgroup$
3
$\begingroup$

This is a special case of Routh's Theorem.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.