# Look for a clever geometric approach to solve a problem with area.

In the acute $$\triangle ABC$$, the position of each point is as shown, $$F$$ is the midpoint of $$BD$$, $$G$$ is the midpoint of $$CE$$, $$F$$ is not on $$CE$$, and $$G$$ is not on $$BD$$. It is known that $$S_{\triangle AFG} = 1$$, and the area of ​​the quadrilateral $$BCDE$$ is obtained?

An imprecise answer: Considering $$D$$, $$E$$ and $$A$$ coincide, it is easy to know that the area of ​​the quadrilateral $$BCDE$$ is $$4$$.

Or:Let $$\overrightarrow{AB}=\vec{a}, \overrightarrow{AC}= \vec{b}, \overrightarrow{AE}=\lambda_1 \vec{a}, \overrightarrow{AD}=\lambda_2 \vec{b}, (0 < \lambda_1, \lambda_2<1)$$, then \begin{aligned} S_{\triangle ABC} &= \frac12|\vec{a}\times\vec{b}| \\ S_{\triangle AED} &=\frac12 \lambda_1 \lambda_2|\vec{a} \times \vec{b}| \\ S_{BCDE} &=S_{\Delta A B C}-S_{\Delta A E D} \\ &=\frac{1}{2}\left(1-\lambda_{1} \lambda_{2}\right)|\vec{a} \times \vec{b}| \end{aligned} $$\because$$ $$\overrightarrow{AF}=\frac{\vec{a}+\lambda_2 \vec{b}}{2} \quad \overrightarrow{A G}=\frac{\lambda_{1} \vec{a}+\vec{b}}{2}$$

$$\therefore$$

\begin{aligned} S_{\Delta A F G} &=\frac{1}{2}\left|\overrightarrow{A F} \times \overrightarrow{A G}\right| \\ &=\frac{1}{8}\left|\left(\vec{a}+\lambda_{2} \vec{b}\right) \times\left(\lambda_{1} \vec{a}+\vec{b}\right)\right| \\ &=\frac{1}{8}\left(1-\lambda_{1} \lambda_{2}\right)|\vec{a} \times \vec{b}| \end{aligned}

$$\therefore$$ $$\frac{S_{\triangle AFG}}{S_{BCDE}}=\frac{1}{4} \Longrightarrow S_{BCDE}=4$$

Now, I need a purely geometric approach to solve this problem, but I have no clue. Could anyone help me? Thanks a lot.

Let $$P$$ be the midpoint of $$DE.$$ Show that $$PF \| AB$$ and $$PG \| AC.$$ Show that $$P$$ lies inside $$\triangle AFG,$$ and therefore $$S_{\triangle AFG} = S_{\triangle AFP} + S_{\triangle AGP} + S_{\triangle PFG}.$$ Show that $$S_{\triangle AFP} = S_{\triangle EFP}$$ and $$S_{\triangle AGP} = S_{\triangle DGP}$$. Show that the area of the quadrilateral $$DEFG$$ is $$S_{DEFG} = S_{\triangle EFP} + S_{\triangle PFG} + S_{\triangle DGP}$$ and therefore $$S_{\triangle AFG} = S_{DEFG}.$$
Now show that $$S_{DEFG} = S_{\triangle DEF} + S_{\triangle DFG} = \tfrac12 S_{\triangle DEB} + \tfrac12 S_{\triangle DBG} = \tfrac12 S_{DEBG}$$ and that $$S_{DEBG} = S_{\triangle DEG} + S_{\triangle BEG} = \tfrac12 S_{\triangle DEC} + \tfrac12 S_{\triangle BEC} = \tfrac12 S_{BCDE}.$$ Therefore $$S_{DEFG} = \frac14 S_{BCDE}.$$