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Given $\triangle ABC$, locate points $A_1$, $B_1$, $C_1$ on respective sides $BC$, $CA$, $AB$ such that $$\frac{BA_1}{A_1C} =\frac{CB_1}{B_1A} = \frac{AC_1}{C_1B} = 2$$

How can I show that the area of the triangle formed by the intersections of $AA_1$, $BB_1$, and $CC_1$ is $1/7$ of the area of $\triangle ABC$?

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Observe how this whole problem is invariant under affine transformation. And any triangle can be mapped to any other via affine transformation. So if you have demonstrated the claim for one triangle, it must be true for all. Pick a particularly simple one. The numbers $3$ and $7$ seem to play a major role, so I chose two edge lengths to be $21$ and it works out nicely: all points are at integer coordinates, making the computation really easy:

Figure

The corners of the inner triangle are at

$$A_2=(12,3)\qquad B_2=(6,12)\qquad C_2=(3,6)$$

so its area is

$$\frac12\begin{vmatrix}12&6&3\\3&12&6\\1&1&1\end{vmatrix}=\frac{63}{2}$$

Compare that to the area of the whole triangle, and you get

$$\frac{63}{21^2}=\frac{63}{441}=\frac17$$

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