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} $\because$ $$\overrightarrow{AF}=\frac{\vec{a}+\lambda_2 \vec{b}}{2} \quad \overrightarrow{A G}=\frac{\lambda_{1} \vec{a}+\vec{b}}{2}$$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.