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In $\triangle ABC$ AA1 and CC1 are medians. On AC is chosen random point P, and then through P are drawn 2 parallel lines lines, one for AA1 and other one for CC1. These 2 lines then intersect AB and BC in F and E. Prove that AA1 and CC1 divide EF in three parts of the same length. The solution which was given used vectors in order to solve this, I was wondering if there is some nice elegant way to prove this without the usage of vectors.enter image description here

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Let $CC_1$ intersect $AA_1$, $FE$ and $PE$ at $G$, $X$ and $Y$, respectively. Observe that $$\frac 13 = \frac{A_1G}{AA_1} = \frac{EY}{PE} = \frac{EX}{FE},$$ so $EX=\frac 13 EF$.

Denoting the intersection of $AA_1$ and $EF$ by $Z$ we find analogously $FZ=\frac 13 EF$. Therefore $XZ=\frac 13 EF$. The conclusion follows.

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  • $\begingroup$ Why does 2nd equality hold? $\endgroup$ – Dovla Feb 14 '17 at 19:36
  • $\begingroup$ This is because $PE \parallel AA_1$ $\endgroup$ – timon92 Feb 14 '17 at 20:47

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