Find the area of a triangle whose vertices cut the sides of $ABC$ in thirds

Question

I have to find the area of $F$, given the following configuration:

$\hspace1in$ What to do?

Answer 1

Hint

I've introduced some notations on your figure:

The angles can be calculated by the law of cosine. For example, for $\alpha$ we have

$$27^2=30^2+51^2-2\cdot30\cdot 51\cdot \cos(\alpha).$$

Having calculated the angles, we can calculate $c,c',c''$ by the law of cosines.

Having calculated these lengths we can calculate one of the angles of the small (shaded) triangle (by the law of cosines).

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Answer 2

Since $BA'=\frac{1}{3}BC$ and so on, we have: $$ [AB'C']=[A'BC']=[A'B'C] = \frac{1}{3}\cdot\frac{2}{3}[ABC] \tag{1}$$ so the area of $F$ is just one third of the area of $ABC$.

The latter can be computed through Heron's formula: $$ [ABC]=\frac{1}{4}\sqrt{108\cdot 6\cdot 48\cdot 54}=324, \tag{2}$$ so:

$$ F = \color{red}{108}.\tag{3}$$