Similarity of nested triangles $ABC$ is a triangle and $A_1, B_1, C_1$ are points on $BC, CA, AB$ such that $$\frac{BA_1}{A_1C}=\frac{CB_1}{B_1A}=\frac{AC_1}{C_1B}=\lambda$$
If $A_2, B_2, C_2$ are points on $B_1C_1, C_1A_1$, and $A_1B_1$ such that $$\frac{B_1A_2}{A_2C_1}=\frac{C_1B_2}{B_2A_1}=\frac{A_1C_2}{C_2B_1}=\frac{1}{\lambda}$$
prove that $\triangle ABC$ is similar to $\triangle A_2B_2C_2$ and find the ratio of simlitude.
I'm missing something obvious, but I don't know what.
 A: $$\vec{A_2B_2}=\vec{A_2C_1}+\vec{C_1B_2}=\frac{1}{1+\frac{1}{\lambda}}\vec{B_1C_1}+\frac{\frac{1}{\lambda}}{1+\frac{1}{\lambda}}\vec{C_1A_2}=$$
$$=\frac{\lambda}{1+\lambda}\vec{B_1C_1}+\frac{1}{1+\lambda}\vec{C_1A_2}=\frac{\lambda}{1+\lambda}\left(\vec{B_1A}+\vec{AC_1}\right)+\frac{1}{1+\lambda}\left(\vec{C_1B}+\vec{BA_1}\right)=$$
$$=\frac{\lambda}{1+\lambda}\left(\frac{1}{1+\lambda}\vec{CA}+\frac{\lambda}{1+\lambda}\vec{AB}\right)+\frac{1}{1+\lambda}\left(\frac{1}{1+\lambda}\vec{AB}+\frac{\lambda}{1+\lambda}\vec{BC}\right)=$$
$$=\frac{\lambda}{(1+\lambda)^2}\left(\vec{BC}+\vec{CA}\right)+\frac{1+\lambda^2}{(1+\lambda)^2}\vec{AB}=\frac{1-\lambda+\lambda^2}{(1+\lambda)^2}\vec{AB},$$
which says $A_2B_2||AB.$
By the same way we can prove that $A_2C_2||AC$ and $B_2C_2||BC,$
which says that $$\Delta ABC\sim\Delta A_2B_2C_2$$ and
$$\frac{A_2B_2}{AB}=\frac{1-\lambda+\lambda^2}{(1+\lambda)^2}.$$
A: A more thorough hint: 
Use the theorem of sines on triangles $BB_2C_1$ and $BB_2A_1$, to conclude that:
$$\dfrac{\sin\angle B_2BA}{\sin\angle B_2BC} = \dfrac{A_1C}{BC_1}.$$
Then the sine version of Ceva's theorem will let you conclude that $AA_2, BB_2, CC_2$ are concurrent. This will greatly help you prove the $A_2B_2//AB$, for example.
Let me know if you still have trouble after this. 
A: We have $B_1 = \frac{1}{1+\lambda} A +\frac{\lambda}{1+\lambda} C$ and cyclic identities.
We have $B_2 = \frac{\lambda}{1+\lambda} A_1 +\frac{1}{1+\lambda} C_1$ and cyclic identities.
By combining them
$$ B_2 = \frac{\lambda}{1+\lambda}\left(\frac{1}{1+\lambda}C+\frac{\lambda}{1+\lambda}B\right)+\frac{1}{1+\lambda}\left(\frac{1}{1+\lambda}B+\frac{\lambda}{1+\lambda}A\right)$$
$$ B_2 = \frac{\lambda}{(1+\lambda)^2}(A+B+C)+\color{red}{\frac{\lambda^2-\lambda+1}{(\lambda+1)^2}} B $$
where the red term is the wanted ratio of similitude.
