6. 如图,在$□ ABCD$中,$AC,BD$相交于点$O$,点$E,F$在$AC$上,$AE=CF$.
(1)求证:四边形$EBFD$是平行四边形;
(2)若$∠ BAC=∠ DAC$,求证:四边形$EBFD$是菱形.

6. (1)证明:$\because$四边形$ABCD$为平行四边形,
$\therefore AO=CO,BO=DO$.
$\because AE=CF$,
$\therefore AO-AE=CO-CF$,
即$EO=FO$,
$\therefore$四边形$EBFD$是平行四边形.
(2)证明:$\because$四边形$ABCD$为平行四边形,
$\therefore AB// CD,\therefore ∠ BAC=∠ DCA$.
又$\because ∠ BAC=∠ DAC,\therefore ∠ DCA=∠ DAC,\therefore DA=DC$.
$\therefore □ ABCD$是菱形.
$\therefore AC⊥ BD$,
即$EF⊥ BD$.
由(1)知四边形$EBFD$是平行四边形,
$\therefore$四边形$EBFD$是菱形.

7. 如图,在$□ ABCD$中,$E,F$分别为$AB,CD$的中点,$BD$是对角线,$AG// DB$,交$CB$的延长线于点$G$,连接$GF$,若$AD⊥ BD$,给出下列结论:①$DE// BF$;②四边形$BEDF$是菱形;③$FG⊥ AB$;④$S_{△ BFG}=\dfrac{1}{4}S_{□ ABCD}$.其中,正确的是()

A. ①②③④
B. ①②
C. ①③
D. ①②④

7. D

8. 如图,在四边形$ABCD$中,$E,F$分别是线段$AD,BC$的中点,$G,H$分别是线段$BD,AC$的中点,当四边形$ABCD$的边满足时,四边形$EGFH$是菱形.

8. $AB=CD$

9. 如图,在矩形$ABCD$中,点$E$为$AB$上任意一点,连接$CE$,点$F$为线段$CE$的中点,过点$F$作$MN⊥ CE$,$MN$与$AB,CD$分别相交于点$M,N$.连接$CM,EN$.
(1)求证:四边形$CNEM$为菱形;
(2)若$AB=10$,$AD=4$,当$AE=2$时,求$EM$的长.

9. (1)证明:在矩形$ABCD$中,$AB// DC$,
$\therefore ∠ MEF=∠ NCF,∠ EMF=∠ CNF$.
又$\because$点$F$为$CE$的中点,
$\therefore EF =CF$.
$\therefore △ EFM≌△ CFN$,
$\therefore EM=CN$.
$\therefore$四边形$CNEM$为平行四边形.
$\because MN⊥ CE$,
$\therefore$四边形$CNEM$为菱形.
(2)解:在菱形$CNEM$中,设$ME=MC=x$.
$\because AB=10,AE=2$,
$\therefore BM=10-2-x=8-x$.
$\because$矩形$ABCD$中,$∠ B=90°,BC=4$,
$\therefore MC^{2}=MB^{2}+BC^{2},\therefore x^{2}=(8-x)^{2}+4^{2}$.
$\therefore x=5$.
即$EM=5$.