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(I have re-labelled the coordinates of the inner triangle, which becomes PQR instead of DEF. Also, L, M and N are the points of trisection on BC, CA and AB.)

The answer is that the area of PQR is one-seventh of the area of ABC.

The only way I can see to prove this is by using vectors. Leta,bandcbe position vectors of the points A, B and C. The point L has position vector (1/3)(2b+c), and similarly (1/3)(2c+a) for M, and (1/3)(2a+b) for N.

The point P lies on both BM and CN, so its position vectorpmust be of the form and also of the form . Solve these equations to see that . Similarly, the position vectors of Q and R are and .

Now think of all these vectors as being in 3-dimensional space (with z-coordinate 0). The standard formula for the area of a triangle is given by a cross product in the form area(ABC) = . If you apply the same formula to the triangle PQR then you get its area as (using the above formulas forp,qandrand simplifying the result. This is 1/7th of the area of ABC.