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1. 如图,在四边形ABCD中,$AD// BC,AD= 5cm,BC= 10cm$,点P从点A出发,以1 cm/s的速度向点D运动,同时,点Q从点C以相同的速度向点B运动.当点P运动到点D时,点Q也随之停止运动.若设运动的时间为t s,以点A,B,C,D,P,Q任意四个点为顶点的四边形中同时存在两个平行四边形,则t的值是 ( )
A.2
B.3
C.4
D.5
A.2
B.3
C.4
D.5
答案：
D
2. 在矩形ABCD中,$AB= 3,BC= 4$,E,F是对角线AC上的两个动点,分别从A,C同时出发相向而行,速度均为1 cm/s,运动时间为t s,$0≤t≤5$.
(1)用含有t的代数式表示EF的长.
(2)若G,H分别是AB,DC中点,连接GE,GF,HE,HF,求证:四边形EGFH是平行四边形.
(3)在(2)的条件下,直接写出当t为何值时,四边形EGFH为矩形.
(1)用含有t的代数式表示EF的长.
(2)若G,H分别是AB,DC中点,连接GE,GF,HE,HF,求证:四边形EGFH是平行四边形.
(3)在(2)的条件下,直接写出当t为何值时,四边形EGFH为矩形.
答案：
(1)解:
∵四边形 ABCD 是矩形,$AB=3,BC=4,\therefore ∠B=90^{\circ },\therefore AC=\sqrt {AB^{2}+BC^{2}}=\sqrt {3^{2}+4^{2}}=5$,由题意,得$AE=CF=t$,E,F 相遇时,即$AE+CF=5,\therefore t+t=5,\therefore t=2.5$,
∴ EF 相遇前$EF=AC - AE - CF=5 - 2t(0≤t＜2.5)$;EF 相遇后$EF=AE + CF - AC=2t - 5(2.5＜t≤5)$. 综上,$0≤t＜2.5$时,EF 的长为$5 - 2t,2.5＜t≤5$时,EF 的长为$2t - 5$.
(2)证明:
∵四边形 ABCD 是矩形,
∴$AB = CD$,$AB// CD$,
∴$∠GAF = ∠HCE$.
∵G,H 分别是 AB,DC 的中点,
∴$AG = BG=\frac{1}{2}AB$,$CH = DH=\frac{1}{2}CD$,
∴$AG = CH$.
∵$AE = CF$,
∴$AF = CE$,
∴$△AFG≌△CEH(SAS)$,
∴$GF = HE$.同理可得$△AGE≌△CHF$,
∴$GE = HF$,
∴四边形 EGFH 是平行四边形.
(3)解:$t = 0.5$或 4.5. 如图,连接 GH. 由
(2)可知,$BG = CH$.
∵$AB// CD$,$∠B = 90^{\circ }$,
∴四边形 GBCH 是矩形,
∴$GH = BC = 4$,
∵四边形 EGFH 是平行四边形,
∴当$EF = GH = 4$时,四边形 EGFH 是矩形.分两种情况:①$0≤t＜2.5$时,$EF = 5 - 2t = 4$,解得$t = 0.5$;②$2.5＜t≤5$时,$EF = 2t - 5 = 4$,解得$t = 4.5$. 综上,当$t = 0.5$或 4.5 时,四边形 EGFH 为矩形.
(1)解:
∵四边形 ABCD 是矩形,$AB=3,BC=4,\therefore ∠B=90^{\circ },\therefore AC=\sqrt {AB^{2}+BC^{2}}=\sqrt {3^{2}+4^{2}}=5$,由题意,得$AE=CF=t$,E,F 相遇时,即$AE+CF=5,\therefore t+t=5,\therefore t=2.5$,
∴ EF 相遇前$EF=AC - AE - CF=5 - 2t(0≤t＜2.5)$;EF 相遇后$EF=AE + CF - AC=2t - 5(2.5＜t≤5)$. 综上,$0≤t＜2.5$时,EF 的长为$5 - 2t,2.5＜t≤5$时,EF 的长为$2t - 5$.
(2)证明:
∵四边形 ABCD 是矩形,
∴$AB = CD$,$AB// CD$,
∴$∠GAF = ∠HCE$.
∵G,H 分别是 AB,DC 的中点,
∴$AG = BG=\frac{1}{2}AB$,$CH = DH=\frac{1}{2}CD$,
∴$AG = CH$.
∵$AE = CF$,
∴$AF = CE$,
∴$△AFG≌△CEH(SAS)$,
∴$GF = HE$.同理可得$△AGE≌△CHF$,
∴$GE = HF$,
∴四边形 EGFH 是平行四边形.
(3)解:$t = 0.5$或 4.5. 如图,连接 GH. 由
(2)可知,$BG = CH$.
∵$AB// CD$,$∠B = 90^{\circ }$,
∴四边形 GBCH 是矩形,
∴$GH = BC = 4$,
∵四边形 EGFH 是平行四边形,
∴当$EF = GH = 4$时,四边形 EGFH 是矩形.分两种情况:①$0≤t＜2.5$时,$EF = 5 - 2t = 4$,解得$t = 0.5$;②$2.5＜t≤5$时,$EF = 2t - 5 = 4$,解得$t = 4.5$. 综上,当$t = 0.5$或 4.5 时,四边形 EGFH 为矩形.
3. 如图所示,在菱形ABCD中,$AB= 4,∠BAD= 120^{\circ },△AEF$为正三角形,点E,F分别在菱形的边BC,CD上滑动,且点E,F不与点B,C,D重合.
(1)证明:不论点E,F在边BC,CD上如何滑动,总有$BE= CF;$
(2)当点E,F在边BC,CD上滑动时,四边形AECF的面积是否发生变化? 如果不变,求出四边形AECF的面积;如果变化,请说明理由.
(1)证明:不论点E,F在边BC,CD上如何滑动,总有$BE= CF;$
(2)当点E,F在边BC,CD上滑动时,四边形AECF的面积是否发生变化? 如果不变,求出四边形AECF的面积;如果变化,请说明理由.
答案：
(1)证明:连接 AC,如图所示.在菱形 ABCD 中,$AB = BC$,$∠BAC = ∠DAC=\frac{1}{2}∠BAD = 60^{\circ }$,
∴$△ABC$是等边三角形,$∠1 + ∠EAC = 60^{\circ }$,
∴$AB = AC$,$∠B = 60^{\circ }$,$\because △AEF$为正三角形,
∴$∠3 + ∠EAC = 60^{\circ }$,
∴$∠1 = ∠3$,$\because AB// CD$,
∴$∠4 = ∠BAC = 60^{\circ }$. 在$△ABE$和$△ACF$中,$\begin{cases} ∠1 = ∠3 \\ AB = AC \\ ∠ABC = ∠4 \end{cases}$,$\therefore △ABE≌△ACF(ASA)$,$\therefore BE = CF$.
(2)解:四边形 AECF 的面积不变. 理由:由
(1)得$△ABE≌△ACF$,则$S_{△ABE}=S_{△ACF}$,$\therefore S_{四边形AECF}=S_{△AEC}+S_{△ACF}=S_{△AEC}+S_{△ABE}=S_{△ABC}$.
∵$△ABC$是等边三角形,$AB = 4$,
∴$S_{△ABC}$是定值. 如图,作$AH⊥BC$于点 H.$\because ∠B = 60^{\circ }$,$\therefore ∠BAH = 30^{\circ }$,$\therefore BH=\frac{1}{2}AB = 2$,$\therefore S_{四边形AECF}=S_{△ABC}=\frac{1}{2}BC\cdot AH=\frac{1}{2}BC\cdot \sqrt{AB^{2}-BH^{2}}=4\sqrt{3}$.
(1)证明:连接 AC,如图所示.在菱形 ABCD 中,$AB = BC$,$∠BAC = ∠DAC=\frac{1}{2}∠BAD = 60^{\circ }$,
∴$△ABC$是等边三角形,$∠1 + ∠EAC = 60^{\circ }$,
∴$AB = AC$,$∠B = 60^{\circ }$,$\because △AEF$为正三角形,
∴$∠3 + ∠EAC = 60^{\circ }$,
∴$∠1 = ∠3$,$\because AB// CD$,
∴$∠4 = ∠BAC = 60^{\circ }$. 在$△ABE$和$△ACF$中,$\begin{cases} ∠1 = ∠3 \\ AB = AC \\ ∠ABC = ∠4 \end{cases}$,$\therefore △ABE≌△ACF(ASA)$,$\therefore BE = CF$.
(2)解:四边形 AECF 的面积不变. 理由:由
(1)得$△ABE≌△ACF$,则$S_{△ABE}=S_{△ACF}$,$\therefore S_{四边形AECF}=S_{△AEC}+S_{△ACF}=S_{△AEC}+S_{△ABE}=S_{△ABC}$.
∵$△ABC$是等边三角形,$AB = 4$,
∴$S_{△ABC}$是定值. 如图,作$AH⊥BC$于点 H.$\because ∠B = 60^{\circ }$,$\therefore ∠BAH = 30^{\circ }$,$\therefore BH=\frac{1}{2}AB = 2$,$\therefore S_{四边形AECF}=S_{△ABC}=\frac{1}{2}BC\cdot AH=\frac{1}{2}BC\cdot \sqrt{AB^{2}-BH^{2}}=4\sqrt{3}$.
4. 点P是线段AB上的动点,分别以AP,BP为边在AB的同侧作正方形APCD与正方形PBEF.
(1)如图(1),连接AF,BC,判断AF与CB的位置关系和数量关系,并证明.
(2)如图(2),将正方形PBEF绕点P逆时针旋转,使得点E落在线段BC上,EF交PC于点G,连接DF,AF,若$DA= DF,AF= 10$,求$S_{△CEG}$.
(3)如图(3),将正方形PBEF绕点P旋转至如图的位置,且$AP= PB$,连接AF,BC,作$∠CPF$的平分线交AF于点H,请写出AH,PH,HF之间的数量关系,并证明.
(1)如图(1),连接AF,BC,判断AF与CB的位置关系和数量关系,并证明.
(2)如图(2),将正方形PBEF绕点P逆时针旋转,使得点E落在线段BC上,EF交PC于点G,连接DF,AF,若$DA= DF,AF= 10$,求$S_{△CEG}$.
(3)如图(3),将正方形PBEF绕点P旋转至如图的位置,且$AP= PB$,连接AF,BC,作$∠CPF$的平分线交AF于点H,请写出AH,PH,HF之间的数量关系,并证明.
答案：
解:
(1)$AF = CB$,$AF⊥CB$. 证明:如图
(1),延长 AF 交 BC 于点 O. 在正方形 APCD 和正方形 PBEF 中,$AP = CP$,$PF = PB$,$∠APF = ∠BPF = 90^{\circ }$. 在$△APF$和$△CPB$中,$\begin{cases} AP = CP \\ ∠APF = ∠CPB \\ PF = PB \end{cases}$,$\therefore △APF≌△CPB(SAS)$,$\therefore AF = CB$,$∠PAF = ∠PCB$.$\because ∠AFP = ∠CFO$,$\therefore ∠APF = ∠COF = 90^{\circ }$,$\therefore AF⊥BC$.
(2)过点 D 作$DN⊥AF$于点 N,如图
(2).$\because DA = DF$,$\therefore AN = NF = 5$. 同
(1)可证$△APF≌△CPB$,$\therefore ∠AFP = ∠B = 90^{\circ }$,$\therefore ∠FAP + ∠APF = 90^{\circ }$.$\because ∠DAN + ∠FAP = 90^{\circ }$,$\therefore ∠DAN = ∠APF$. 又$\because ∠AND = ∠AFP$,$AD = AP$,$\therefore △AND≌△PFA(AAS)$,$\therefore AN = PF = 5$. 由$△AFP≌△CBP$,得$PF = PB = 5$,$AF = BC = 10$.$\because BE = PF = 5$,$\therefore CE = 5$.$\because EG// PB$,$\therefore EG=\frac{1}{2}PB=\frac{5}{2}$,$\therefore S_{△EGC}=\frac{1}{2}CE\cdot EG=\frac{1}{2}×5×\frac{5}{2}=\frac{25}{4}$.
(3)$AH = HF+\sqrt{2}PH$. 证明:如图
(3),在 AH 上截取$AM = HF$,连接 MP.
∵ 在正方形 APCD 和正方形 PBEF 中,$AP = PB$,$\therefore AP = PC = PF$,$\therefore ∠PAM = ∠PFH$. 又$\because AM = HF$,$\therefore △APM≌△FPH(SAS)$,$\therefore PM = PH$,$∠APM = ∠FPH$.$\because PH$平分$∠CPF$,$\therefore ∠CPH = ∠FPH$,$\therefore ∠APM = ∠CPH$.$\because ∠APC = 90^{\circ }$,$\therefore ∠APM + ∠MPC = ∠CPH + ∠MPC = ∠MPH = 90^{\circ }$,$\therefore MH=\sqrt{2}PH$.$\because AH = AM + MH$,$\therefore AH = HF+\sqrt{2}PH$.
解:
(1)$AF = CB$,$AF⊥CB$. 证明:如图
(1),延长 AF 交 BC 于点 O. 在正方形 APCD 和正方形 PBEF 中,$AP = CP$,$PF = PB$,$∠APF = ∠BPF = 90^{\circ }$. 在$△APF$和$△CPB$中,$\begin{cases} AP = CP \\ ∠APF = ∠CPB \\ PF = PB \end{cases}$,$\therefore △APF≌△CPB(SAS)$,$\therefore AF = CB$,$∠PAF = ∠PCB$.$\because ∠AFP = ∠CFO$,$\therefore ∠APF = ∠COF = 90^{\circ }$,$\therefore AF⊥BC$.
(2)过点 D 作$DN⊥AF$于点 N,如图
(2).$\because DA = DF$,$\therefore AN = NF = 5$. 同
(1)可证$△APF≌△CPB$,$\therefore ∠AFP = ∠B = 90^{\circ }$,$\therefore ∠FAP + ∠APF = 90^{\circ }$.$\because ∠DAN + ∠FAP = 90^{\circ }$,$\therefore ∠DAN = ∠APF$. 又$\because ∠AND = ∠AFP$,$AD = AP$,$\therefore △AND≌△PFA(AAS)$,$\therefore AN = PF = 5$. 由$△AFP≌△CBP$,得$PF = PB = 5$,$AF = BC = 10$.$\because BE = PF = 5$,$\therefore CE = 5$.$\because EG// PB$,$\therefore EG=\frac{1}{2}PB=\frac{5}{2}$,$\therefore S_{△EGC}=\frac{1}{2}CE\cdot EG=\frac{1}{2}×5×\frac{5}{2}=\frac{25}{4}$.
(3)$AH = HF+\sqrt{2}PH$. 证明:如图
(3),在 AH 上截取$AM = HF$,连接 MP.
∵ 在正方形 APCD 和正方形 PBEF 中,$AP = PB$,$\therefore AP = PC = PF$,$\therefore ∠PAM = ∠PFH$. 又$\because AM = HF$,$\therefore △APM≌△FPH(SAS)$,$\therefore PM = PH$,$∠APM = ∠FPH$.$\because PH$平分$∠CPF$,$\therefore ∠CPH = ∠FPH$,$\therefore ∠APM = ∠CPH$.$\because ∠APC = 90^{\circ }$,$\therefore ∠APM + ∠MPC = ∠CPH + ∠MPC = ∠MPH = 90^{\circ }$,$\therefore MH=\sqrt{2}PH$.$\because AH = AM + MH$,$\therefore AH = HF+\sqrt{2}PH$.
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