已知平行四边形$ABCD$,点$G$在$AB$上,满足$AG = 2GB$;点$E$在$DC$上,满足$CE = 2DE$;点$F$在$BC$上,满足$BF = 2FC$。求$\triangle EFG$的面积占平行四边形面积的几分之几。
已知
$ABCD$是平行四边形,点$G$在$AB$上,满足$AG = 2GB$;点$E$在$DC$上,满足$CE = 2DE$;点$F$在$BC$上,满足$BF = 2FC$。
求解
我们需要求$\triangle EFG$的面积占平行四边形面积的几分之几。
解答
作$EP \perp AB$和$EQ \perp BC$
$AB=2GB, CE=2DE$且$BF=2FC$
这意味着:
$AB - GB = 2GB$
$CD - DE = 2DE$
$BC - FC = 2FC$
$AB = 3BG$,$CD = 3DE$
$BC = 3FC$
$GB = \frac{1}{3}AB$,$DE = \frac{1}{3}CD$,$FC = \frac{1}{3}BC$.........(i)
$\operatorname{ar}(\triangle EFG) = \operatorname{ar}(\text{梯形BGEC}) - \operatorname{ar}(\triangle BGF)$............(vii)
$=\frac{1}{2}(\frac{1}{3}AB + \frac{2}{3}CD) \times EP$
$=\frac{1}{2}AB \times EP$
$=\frac{1}{2}\operatorname{ar}(\text{平行四边形ABCD})$
$\operatorname{ar}(\triangle EFC) = \frac{1}{9}\operatorname{ar}(\text{平行四边形ABCD})$
$\operatorname{ar}(\triangle BGF) = \frac{1}{2}BF \times GR$
$=\frac{1}{2} \times \frac{2}{3}BC \times GR$
$=\frac{2}{3} \times \frac{1}{2}BC \times GR$
$=\frac{2}{3} \times \operatorname{ar}(\triangle GBC)$
$=\frac{2}{3} \times \frac{1}{2} GB \times EP$
$=\frac{1}{3} \times \frac{1}{3}AB \times EP$
$=\frac{1}{9}AB \times EP$
$=\frac{1}{9}\operatorname{ar}(\text{平行四边形ABCD})$
$\operatorname{ar}(\triangle EFG) = \frac{1}{2}\operatorname{ar}(\text{平行四边形ABCD}) - \frac{1}{9}\operatorname{ar}(\text{平行四边形ABCD}) - \frac{1}{9}\operatorname{ar}(\text{平行四边形ABCD}) = \frac{5}{18}\operatorname{ar}(\text{平行四边形ABCD})$