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1. 如图,将平行四边形$ABCD$绕点$A$逆时针旋转得到平行四边形$AEFG$,使点$E$落在边$BC$上,且$D$恰好是$FG$的中点,若$\frac{AB}{AD}=\frac{4}{5}$,则$\frac{BE}{CE}$的值为
$\frac{8}{17}$
.
答案：
1. $\frac{8}{17}$
2. 如图,$D$是$Rt△ ABC$的斜边$AC$上一点,且$∠ ABC=90^{\circ},∠ A=30^{\circ},BC=4\sqrt{2}$,以$BD$为斜边作等腰$Rt△ BDE$,使点$E,C$在$BD$同侧,连接$CE$,则$CE$的最小值为
2
.
答案：
2. 点拨:如答图,过点C作$CH⊥BC$,使$CH=CB=4\sqrt{2}$,连接BH,HD,则$△ HCB$为等腰直角三角形，
$\therefore ∠HBC=45^{\circ }$.
∵$△ BDE$为等腰直角三角形,
$\therefore ∠DBE=∠BDE=45^{\circ },BE=DE$,
$\therefore ∠DBE=∠HBC,\therefore ∠DBH=∠EBC$.
$\because \frac{BD}{BE}=\frac{HB}{BC}=\sqrt{2},\therefore △ BDH∽ △ BEC$,
$\therefore CE=\frac{\sqrt{2}}{2}DH$,$\therefore$当DH取最小值时,CE最小,
$\therefore$当$HD⊥AC$时,DH最小.$\because CH⊥BC,AB⊥BC$,
$\therefore CH// AB,\therefore ∠HCD=∠A=30^{\circ },\therefore HD=\frac{1}{2}HC=2\sqrt{2}$,$\therefore CE=\frac{\sqrt{2}}{2}DH=2$,$\therefore$ CE的最小值为2.
2. 点拨:如答图,过点C作$CH⊥BC$,使$CH=CB=4\sqrt{2}$,连接BH,HD,则$△ HCB$为等腰直角三角形，
$\therefore ∠HBC=45^{\circ }$.
∵$△ BDE$为等腰直角三角形,
$\therefore ∠DBE=∠BDE=45^{\circ },BE=DE$,
$\therefore ∠DBE=∠HBC,\therefore ∠DBH=∠EBC$.
$\because \frac{BD}{BE}=\frac{HB}{BC}=\sqrt{2},\therefore △ BDH∽ △ BEC$,
$\therefore CE=\frac{\sqrt{2}}{2}DH$,$\therefore$当DH取最小值时,CE最小,
$\therefore$当$HD⊥AC$时,DH最小.$\because CH⊥BC,AB⊥BC$,
$\therefore CH// AB,\therefore ∠HCD=∠A=30^{\circ },\therefore HD=\frac{1}{2}HC=2\sqrt{2}$,$\therefore CE=\frac{\sqrt{2}}{2}DH=2$,$\therefore$ CE的最小值为2.
3. 如图①,矩形$EBGF$和矩形$ABCD$共顶点,且绕着点$B$顺时针旋转,满足$\frac{BC}{AB}=\frac{BG}{BE}=\frac{3}{4}$.
(1)$\frac{DF}{AE}$的比值是否发生变化,若变化,说明理由;若不变,求出相应的值,并说明理由;
(2)如图②,若$F$为$CD$的中点,且$AB=8,AD=6$,连接$CG$,求$△ FCG$的面积.
(1)$\frac{DF}{AE}$的比值是否发生变化,若变化,说明理由;若不变,求出相应的值,并说明理由;
(2)如图②,若$F$为$CD$的中点,且$AB=8,AD=6$,连接$CG$,求$△ FCG$的面积.
答案：
3. 解:
(1)$\frac{DF}{AE}=\frac{5}{4}$.如答图①,连接BD,BF.
$\because$ 四边形ABCD是矩形,$\therefore ∠DAB=90^{\circ },AD=BC$.
$\because BC:AB=3:4,\therefore AD:AB=3:4$,设$AD=3k,AB=4k$,则$BD=5k,\therefore AD:AB:BD=3:4:5$,
同法可证$EF:BE:BF=3:4:5$.$\therefore △ ABD∽ △ EBF$,
$\therefore ∠ABD=∠EBF,\frac{AB}{BE}=\frac{BD}{BF}$,
$\therefore ∠ABE=∠DBF,\frac{AB}{BD}=\frac{BE}{BF},\therefore △ ABE∽ △ DBF$,
$\therefore \frac{DF}{AE}=\frac{DB}{AB}=\frac{5}{4}$.
(2)如答图②,连接BF,AE,过点G作$GT⊥DC$交DC的延长线于点T.
$\because$ 四边形ABCD是矩形,$\therefore AB=CD=8,\therefore DF=CF=4$.
$\because DF:AE=5:4,\therefore AE=\frac{16}{5}$.
$\because ∠ABC=∠EBG=90^{\circ },\therefore ∠ABE=∠CBG$.
$\because \frac{AB}{CB}=\frac{BE}{BG}=\frac{4}{3},\therefore △ ABE∽ △ CBG$,
$\therefore \frac{AE}{CG}=\frac{AB}{BC}=\frac{4}{3},\therefore CG=\frac{12}{5}$.$\because ∠BCF=∠BGF=90^{\circ }$,
$\therefore$ C,F,B,G四点共圆,
$\therefore ∠GCT=∠FBG$.$\because ∠T=∠BGF=90^{\circ }$,
$\therefore △ CTG∽ △ BGF$,
$\therefore CT:GT:CG=BG:GF:BF=3:4:5$,
$\therefore GT=\frac{4}{5}CG=\frac{48}{25}$,
$\therefore △ CFG$的面积为$\frac{1}{2}CF· GT=\frac{1}{2}× 4× \frac{48}{25}=\frac{96}{25}$.
3. 解:
(1)$\frac{DF}{AE}=\frac{5}{4}$.如答图①,连接BD,BF.
$\because$ 四边形ABCD是矩形,$\therefore ∠DAB=90^{\circ },AD=BC$.
$\because BC:AB=3:4,\therefore AD:AB=3:4$,设$AD=3k,AB=4k$,则$BD=5k,\therefore AD:AB:BD=3:4:5$,
同法可证$EF:BE:BF=3:4:5$.$\therefore △ ABD∽ △ EBF$,
$\therefore ∠ABD=∠EBF,\frac{AB}{BE}=\frac{BD}{BF}$,
$\therefore ∠ABE=∠DBF,\frac{AB}{BD}=\frac{BE}{BF},\therefore △ ABE∽ △ DBF$,
$\therefore \frac{DF}{AE}=\frac{DB}{AB}=\frac{5}{4}$.
(2)如答图②,连接BF,AE,过点G作$GT⊥DC$交DC的延长线于点T.
$\because$ 四边形ABCD是矩形,$\therefore AB=CD=8,\therefore DF=CF=4$.
$\because DF:AE=5:4,\therefore AE=\frac{16}{5}$.
$\because ∠ABC=∠EBG=90^{\circ },\therefore ∠ABE=∠CBG$.
$\because \frac{AB}{CB}=\frac{BE}{BG}=\frac{4}{3},\therefore △ ABE∽ △ CBG$,
$\therefore \frac{AE}{CG}=\frac{AB}{BC}=\frac{4}{3},\therefore CG=\frac{12}{5}$.$\because ∠BCF=∠BGF=90^{\circ }$,
$\therefore$ C,F,B,G四点共圆,
$\therefore ∠GCT=∠FBG$.$\because ∠T=∠BGF=90^{\circ }$,
$\therefore △ CTG∽ △ BGF$,
$\therefore CT:GT:CG=BG:GF:BF=3:4:5$,
$\therefore GT=\frac{4}{5}CG=\frac{48}{25}$,
$\therefore △ CFG$的面积为$\frac{1}{2}CF· GT=\frac{1}{2}× 4× \frac{48}{25}=\frac{96}{25}$.
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