$ABC$ is a triangle. $A_1B_1C_1$ is a tiangle inside $ABC$ such that $A_1$ divides $BC$, $B_1$ divdes $CA$ and $C_1$ divides $AB$ in $1:2$ ratio. A further trianle $A_2B_2C_2$ is constructed such that $A_2$ divides $B_1C_1$, $B_2$ divdes $C_1A_1$ and $C_2$ divides $A_1B_1$ in $2:1$ ratio.

How to show that

1) $A_2B_2$ is parallel to $AB$ ?

2)$A_2B_2$ is a thrid of $AB$ ?

Through near accurate constructions, and looking as to what would happen if the above two are true, I feel that most of our work would become simple if we could show that $A_2B_2C_2$ is similar to $ABC$.

How can I proceed?

Edit: @Michael Rozenberg, thank you for the simple vector solution. I am looking for a geometric approach, espectially with similar triangles etc. Thank you Michael nevertheless.

  • $\begingroup$ Just to be clear, I assume the $2:1$ ratio applies to how $A_2$ and $B_2$ divide each of their lines in addition to that of $C_2$ dividing $A_1 B_1$. Please let me know if this is not correct. $\endgroup$ – John Omielan Jan 20 at 7:43
  • $\begingroup$ @JohnOmielan You assume right. $\endgroup$ – Qwerty Jan 20 at 8:26

Let $\vec{AB}=\vec{u}$ and $\vec{AC}=\vec{v}$.

Thus, $$\vec{A_2B_2}=\frac{1}{3}\vec{B_1C_1}+\frac{2}{3}\vec{C_1A_1}=$$ $$=\frac{1}{3}\left(-\frac{2}{3}\vec{v}+\frac{1}{3}\vec{u}\right)+\frac{2}{3}\left(\frac{2}{3}\vec{u}+\frac{1}{3}\left(-\vec{u}+\vec{v}\right)\right)=\frac{1}{3}\vec{u}=\frac{1}{3}\vec{AB},$$ which says $$A_2B_2||AB.$$

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