Find the area of a triangle inside of a triangle.

Question

ABC is a triangle with area 1. AF = AB/3, BE = BC/3, ED = FD. Find the area of the shaded figure.

I tried to solve it a lot, I could not get the solution, so please give a solution with proper description. Thanks in advance.

Answer 1

Let me try. We have $$S_{BEF} = \frac{1}{3}\frac{2}{3} S_{BAC} = \frac{2}{9}S_{BAC}.$$ So $$S_{ACEF} = \frac{7}{9}S_{ABC}.$$

$$S_{AFC} = \frac{1}{3}S_{ABC}, S_{AEC} = \frac{2}{3}S_{ABC}$$ then $$S_{CEF} = \frac{4}{9}S_{ABC}$$

$$S_{CED} = \frac{1}{2}S_{CEF} = \frac{1}{2}(S_{ACEF} - S_{AFC}) = \frac{2}{9}S_{ABC}.$$

$$S_{AFD} = \frac{1}{2}S_{AEF} = \frac{1}{2}(S_{ACEF} - S_{AEC}) = \frac{1}{18}S_{ABC}.$$

Thus, $$S_{ACD} = S_{ACEF} - S_{CED} - S_{AFD} = \frac{1}{2}S_{ABC}.$$

Answer 2

Take $A$ as the origin of vectors and let $\vec{AB} = \mathbf{b}$ and $\vec{AC} = \mathbf{c}$. Then $\vec{AF} = \frac{1}{3}\mathbf{b}$ and $\vec{AE} = \frac{\mathbf{c}+2\mathbf{b}}{3}$. Hence $\vec{AD} = \frac{1}{2}\left(\frac{1}{3}\mathbf{b}+\frac{\mathbf{c}+2\mathbf{b}}{3}\right) = \frac{\mathbf{c}+3\mathbf{b}}{6}$. Hence area of triangle $ADC$ equals $$\frac{1}{2}\left| \vec{AD} \times \vec{AC} \right| = \frac{1}{4}|\mathbf{b}\times\mathbf{c}| $$ and hence equals half the area of the triangle $ABC$.