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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? enter image description here

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.

user
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Zuo
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1 Answers1

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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 $S_{DEFG} = \frac14 S_{BCDE}.$

David K
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