# Second intriangle similar to outer triangle

$$ABC$$ 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.

• 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. – John Omielan Jan 20 at 7:43
• @JohnOmielan You assume right. – 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 $$A_2B_2||AB.$$