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8. 如图,在$△ ABC$中,有菱形$AMPN$,点$M$、$P$、$N$分别在$AB$、$BC$、$AC$上.若$\frac{AM}{MB}=\frac{1}{2}$,则$\frac{BP}{BC}=$
$\frac{2}{3}$
.答案
8. $\frac{2}{3}$
9. 如图，在四边形ABCD中，已知$AD// BC$，AC与BD相交于点O。若$S_{△ AOD}=4$，$S_{△ AOB}=6$，则$S_{△ BOC}=$
9
。答案
9. 9
三、解答题
10. 如图，已知$EF// BC$，$FG// CD$.
求证：$\frac{AE}{AB}=\frac{AG}{AD}$.
10. 如图，已知$EF// BC$，$FG// CD$.
求证：$\frac{AE}{AB}=\frac{AG}{AD}$.
答案
10. $\because EF // BC, \therefore ∠ AEF = ∠ ABC, ∠ AFE = ∠ ACB, \therefore △ AEF ∽ △ ABC, \therefore \frac{AE}{AB}=\frac{AF}{AC}$. 同理:$△ AFG ∽ △ ACD, \therefore \frac{AF}{AC}=\frac{AG}{AD}, \therefore \frac{AE}{AB}=\frac{AG}{AD}$.
11. 如图,在△ABC中,已知DG//EC,EG//BC.
求证:AE²=AB·AD.
求证:AE²=AB·AD.
答案
11. $\because DG // EC, \therefore \frac{AD}{AE}=\frac{AG}{AC}$.
$\because EG// BC, \therefore \frac{AG}{AC}=\frac{AE}{AB}, \therefore \frac{AD}{AE}=\frac{AE}{AB}, \therefore AE^2 = AB · AD$.
$\because EG// BC, \therefore \frac{AG}{AC}=\frac{AE}{AB}, \therefore \frac{AD}{AE}=\frac{AE}{AB}, \therefore AE^2 = AB · AD$.
12. 如图，已知四边形ABCD是平行四边形，E是AB延长线上一点，DE分别交对角线AC于点G，交边BC于点F.
(1)求证：$DG^2 = GF · GE$；
(2)求证：$\frac{GC^2}{GA^2} = \frac{GF}{GE}$.
(1)求证：$DG^2 = GF · GE$；
(2)求证：$\frac{GC^2}{GA^2} = \frac{GF}{GE}$.
答案
12. (1) $\because$ 四边形ABCD是平行四边形, $\therefore AD// BC, AB// CD$. $\because AD // BC, \therefore \frac{DG}{GF}=\frac{AG}{GC}$. $\because AB // CD, \therefore \frac{AG}{GC}=\frac{GE}{DG}, \therefore \frac{DG}{GF}=\frac{GE}{DG}, \therefore DG^2 = GF · GE$.
(2) $\because AB// CD, \therefore \frac{GC}{GA}=\frac{DG}{GE}$,
$\therefore \frac{GC^2}{GA^2}=\frac{DG^2}{GE^2}=\frac{GF · GE}{GE^2}=\frac{GF}{GE}$.
(2) $\because AB// CD, \therefore \frac{GC}{GA}=\frac{DG}{GE}$,
$\therefore \frac{GC^2}{GA^2}=\frac{DG^2}{GE^2}=\frac{GF · GE}{GE^2}=\frac{GF}{GE}$.
