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$CP$,$AM$ and $BN$ are medians of triangle $ABC$.These three medians are concurrent in point $G$.A line that passes point $G$ and is parallel to $BC$ intersects $AB$ in $A_b$ and $AC$ in $A_c$.A line that passes point $G$ and is parallel to $AC$ intersects $AB$ in $B_a$ and $BC$ in $B_c$.A line that passes point $G$ and is parallel to $AB$ intersects $AC$ in $C_a$ and $BC$ in $C_b$.Prove triangle $A_cB_aC_b$ and $A_bB_cC_a$ are congruent.

Additional info:We are not allowed to use Ceva.

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My Ideas

If i show that $A_bGCB_c$ and $B_aA_cCG$ are parallelogram,then $B_a =A_c$.similar to this way I can show other sides are equal to each other.So triangles will be congruent.

However, I was just able to observe $A_bGC_bB$ , $GA_cCB_c$ , $AC_aGB_a$ are parallelogram.

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Since $\overline{AM}$ is a median and $\overline{BC}\parallel\overline{A_bA_c}$, $\overline{A_bG}=\overline{GA_c}$. Similarly, $\overline{B_aG}=\overline{GB_c}$. This implies that $\square A_bB_aA_cB_c$ is a parallelogram, so we have $\overline{A_bB_c}=\overline{B_aA_c}$.

In the same manner, $\overline{B_cC_a}=\overline{C_bB_a}$, $\overline{C_aA_b}=\overline{A_cC_b}$. Thus we get the result.

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  • $\begingroup$ thanks for hint.in Addition The reason of $\overline{A_bG}=\overline{GA_c}$,is $\frac{ \overline{A_bG}}{\overline{BM}}=\frac{ \overline{G_Ac}}{\overline{CM}}=\frac{ \overline{AG}}{\overline{AM}}$,And we know that $\overline{BM}=\overline{CM}$.So $\overline{A_bG}=\overline{GA_c}$ is true $\endgroup$ – user2838619 Aug 3 '14 at 13:05

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