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In $\triangle ABC$, $A_1, B_1, C_1$ are points on $BC,CA,AB$ such that $$\frac{BA_1}{A_1C} = \frac{CB_1}{B_1A} = \frac{AC_1}{C_1B} = \delta$$ If $A_2,B_2,C_2$ are points on $B_1C_1, C_1A_1, A_1B_1$ such that $$\frac{B_1A_2}{A_2C_1} = \frac{C_1B_2}{B_2A_1} = \frac{A_1C_2}{C_2B_1} = \frac{1}{\delta}$$ then prove that $\triangle ABC$ is similar to $\triangle A_2B_2C_2$ and find the ratio of similitude.

enter image description here After making a diagram it is pretty much clear that $A_2C_2 || AC, C_2B_2 || BC$ and $B_2A_2 || BA$. Proving this would be sufficient to prove the similarity.

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Working with vectors we have $$B-A_1 = \delta(A_1-C)$$ so $$A_1 = \frac{B+\delta C}{1+\delta}\\ B_1=\frac{C+\delta A}{1+\delta}\\ C_1= \frac{A+\delta B}{1+\delta} $$ Similarly $$A_2=\frac{\delta B_1 + C_1}{1+\delta}=\frac{\delta\frac{C+\delta A}{1+\delta}+\frac{A+\delta B}{1+\delta}}{1+\delta}=\frac{(\delta^2+1)A + \delta B+\delta C}{(1+\delta)^2}\\ B_2=\frac{(\delta^2+1)B + \delta C+\delta A}{(1+\delta)^2}\\ C_2=\frac{(\delta^2+1)C + \delta A+\delta B}{(1+\delta)^2}$$

Therefore $$A_2B_2=\frac{\delta^2+1-\delta}{(1+\delta)^2} AB$$ and the others, so the triangles $ABC$, $A_2B_2C_2$ are similar.

We can rewrite the above equalities as $$A_2= \frac{\delta^2-\delta+1}{(\delta+1)^2}A+\frac{3 \delta}{(\delta+1)^2}\frac{A+B+C}{3}$$

Therefore, the triangle $A_2B_2C_2$ is obtained from $ABC$ with a homothety with center the center of mass of $ABC$ and constant $\frac{\delta^2-\delta+1}{(\delta+1)^2}$

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