《2025年多维导学案八年级数学上册人教版》

(1)若设$S_{\triangle GCD}=x,S_{\triangle GBF}=y,S_{\triangle GAE}=z$猜想x,y,z之间的数量关系,并证明你的猜想;
解:(1)$x=y=z$.证明如下:由题意,得$S_{\triangle GCD}=S_{\triangle GBD}=x$,$S_{\triangle GBF}=S_{\triangle AGF}=y$,$S_{\triangle GAE}=S_{\triangle GCE}=z$.$\because S_{\triangle ABD}=S_{\triangle ACD}$,$\therefore 2y+x=2z+x$.$\therefore y=z$.$\because S_{\triangle ABE}=S_{\triangle CBE}$,$\therefore 2y+z=2x+z$.$\therefore x=y$.$\therefore x=y=z$.
(2)由(1)可知被三条中线分成的六个三角形面积相等,若$\triangle ABC$的面积为m,则用含有m的式子表示$\triangle BGC$的面积为$\frac{1}{3}m$,$BG:GE=$$2:1$;
(3)如图4,点D,E在$\triangle ABC$的边AC,AB上,BD,CE交于点G,点G是$\triangle ABC$的重心,且$BD=6,CE=9,BD⊥CE$,求四边形 AEGD 的面积.
解:(3)$\because$点$G$是$\triangle ABC$的重心,$\therefore BG:GD=CG:GE=2:1$.$\because BD=6$,$CE=9$,$\therefore BG=4$,$CG=6$.$\because BD\perp CE$,$\therefore S_{\triangle BGC}=\frac{1}{2}BG\cdot CG=\frac{1}{2}×4×6=12$.$\therefore S_{\triangle ABC}=3S_{\triangle BGC}=36$,$S_{\triangle BEG}=S_{\triangle CDG}=\frac{1}{2}S_{\triangle BGC}=6$.$\therefore S_{四边形AEGD}=36-6-6-12=12$.

答案： 解:
(1)$x=y=z$.证明如下:
由题意,得$S_{\triangle GCD}=S_{\triangle GBD}=x$,$S_{\triangle GBF}=S_{\triangle AGF}=y$,$S_{\triangle GAE}=S_{\triangle GCE}=z$.
$\because S_{\triangle ABD}=S_{\triangle ACD}$,
$\therefore 2y+x=2z+x$.
$\therefore y=z$.
$\because S_{\triangle ABE}=S_{\triangle CBE}$,
$\therefore 2y+z=2x+z$.
$\therefore x=y$.
$\therefore x=y=z$.
(2)相等 $\frac{1}{3}m$ $2:1$
(3)$\because$点$G$是$\triangle ABC$的重心,
$\therefore BG:GD=CG:GE=2:1$.
$\because BD=6$,$CE=9$,
$\therefore BG=4$,$CG=6$.
$\because BD\perp CE$,
$\therefore S_{\triangle BGC}=\frac{1}{2}BG\cdot CG=\frac{1}{2}\times4\times6=12$.
$\therefore S_{\triangle ABC}=3S_{\triangle BGC}=36$,$S_{\triangle BEG}=S_{\triangle CDG}=\frac{1}{2}S_{\triangle BGC}=6$.
$\therefore S_{四边形AEGD}=36-6-6-12=12$.