矩形的判定专项练习30题
矩形的判定专项练习30题（有答案）
1．如图，在四边形ABCD中，AD∥BC，E、F为AB上两点，且△DAF≌△CBE．
求证：（1）∠A=90°；
（2）四边形ABCD是矩形．
2．如图，已知平行四边形ABCD，∠ABC，∠BCD的平分线BE、CF分别交AD于E、F，BE、CF 交于点G，点H为BC的中点，GH的延长线交GB的平行线CM于点M．
（1）试说明：∠BGC=90°；
（2）连接BM，判断四边形GBMC的形状并说明理由．
3．如图，O是菱形ABCD对角线的交点，作DE∥AC，CE∥BD，DE、CE交于点E．
（1）四边形OCDE是矩形吗？说说你的理由；
（2）请你将上述条件中的菱形改为另一种四边形，其它条件都不变，你能得出什么结论？根据改编后的题目画出图形，并说明理由．
4．△ABC中，AD⊥BC于D，点E、F分别是△ABC中AB、AC中点，当△ABC满足什么条件时，四边形AEDF是矩形？说明理由．
5．如图，菱形ABCD的对角线AC、BD交于点O．
（1）用尺规作图的方法，作出△AOB平移后的△DEC，其中平移的方向为射线AD的方向，平移的距离为线段AD的长；（要求：用尺规作图，保留作图痕迹，不写作法．）
（2）观察图形，判断四边形DOCE是什么特殊四边形，并证明．
6．如图，平行四边形ABCD中，对角线AC、BD相交于点O，延长OA到N，ON=OB，再延长OC 至M，使CM=AN，求证：四边形NDMB为矩形．
7．如图，点O是菱形ABCD对角线的交点，过点C作BD的平行线CE，过点D作AC的平行线DE，CE与DE相交于点E，试说明四边形OCED是矩形．
8．如图，已知梯形ABCD中，AD∥BC，AB⊥BC，点E、F分别是边BC、CD的中点，直线EF交边AD的延长线于点M，连接BD．
（1）求证：四边形DBEM是平行四边形；
（2）若BD=DC，连接CM，求证：四边形ABCM为矩形．
9．如图，在△ABC中，点O是AC边上的中点，过点O的直线MN∥BC，且MN交∠ACB的平分线于点E，交∠ACB的外角平分线于点F，点P是BC延长线上一点．求证：四边形AECF是矩形．
10．如图，在梯形ABCD中，AD∥BC，BC=2AD，点E是BC的中点，连接AC、DE相交于点O．（1）试说明：△AOD≌△COE；
（2）若∠B=∠AOE，试说明四边形AECD是矩形的理由．
11．如图，以△ABC的各边为一边向BC的同侧作正△ABD、正△BCF、正△ACE，若
∠BAC=150°，求证：四边形AEFD为矩形．
12．（1）在等腰三角形ABC中AB=BC，∠ABC=90°，BD⊥AC，过D点作DE⊥DF，交AB于E，交BC于F．若AE=4，FC=3，求EF长．
（2）如图，将?ABCD的边DC延长到点E，使CE=DC，连接AE，交BC于点F．
①求证：△ABF≌△ECF；
②若∠AFC=2∠D，连接AC、BE．求证：四边形ABEC是矩形．
13．如图，AD是△ABC的中线，过点A作AE∥BC，过点B作BE∥AD交AE于点E，
（1）求证：AE=CD；
（2）当△ABC满足什么条件时，四边形ADBE是矩形？请说明理由．
14．如图，已知梯形ABCD中，AD∥BC，E、F分别是AB、CD的中点，点G在边BC上，且CG=（AD+BC）．
（1）求证：四边形DEGF是平行四边形；
（2）连接DG，若∠ADG=2∠ADE，求证：四边形DEGF是矩形．
15．已知，如图在△ABC中，AB=AC，点D是AC的中点，直线AE∥BC，过D点作直线EF∥AB 分别交AE、BC于点E、F，求证：四边形AECF是矩形．
16．已知：如图，在△ABC中，D、E、F分别是AC、AB、BC的中点，且CE=AB．
求证：四边形CFED是矩形．
17．如图，平行四边形ABCD中，EF过AC的中点O，与边AD、BC分别相交于点E、F；（1）试说明四边形AECF是平行四边形．
（2）若EF过AC的中点，且与AC垂直时，试说明四边形AECF是菱形．
（3）当EF与AC有怎样的关系时，四边形AECF是矩形．
18．如图，在Rt△ABC中，∠A=90°，AB=AC，D是斜边BC上一点，DE⊥AC，DF⊥AB，垂足分别为E、F．
（1）说明四边形AEDF是矩形．
（2）试问：当点D位于BC边的什么位置时，四边形AEDF是正方形？并说明你的理由．
19．如图，△ABC中，D为边AC的中点，过点D作MN∥BC，CE平分∠ACB交MN于E，CF平分∠ACG交MN于F，求证：（1）ED=DF；（2）四边形AECF为矩形．
20．如图，菱形ABCD的对角线AC、BC相交于点O，BE∥AC，CE∥DB．求证：四边形OBEC是矩形．
21．如图，在△ABC中，O是AC上的任意一点，（不与点A，C重合），过点O作直线l∥BC，直线l与∠BCA的平分线相交于点E，与∠DCA的平分线相交于点F．
（1）OE与OF相等吗？为什么？
（2）探索：当点O在何处时，四边形AECF为矩形？为什么？
22．（2013?沙湾区模拟）如图，在△ABC中，D是BC边上的一点，E是AD的中点，过A点作BC的平行线交CE的延长线于F，且
AF=BD，连接BF．
（1）求证：D是BC的中点．
（2）如果AB=AC，试判断四边形AFBD的形状，并证明你的结论．
23．如图，四边形ABCD是平行四边形，对角线AC、BD交于点O，∠OBC=∠OCB，求证：四边形ABCD是矩形．
24．如图M、N分别是平行四边形ABCD的对边AD、BC的中点，且AD=2AB，AN，BM相交于P，DN，CM相交于Q．求证：PMQN为矩形．
25．在平行四边形ABCD中，对角线AC、BD相交于O，EF过点O，且AF⊥BC，求证：四边形
AFCE是矩形．
26．如图，在△ABC中，D是AC的中点，E是线段BC延长线上的一点，过点A作AF∥BE，交ED的延长线于点F，连接AE，CF．
（1）求证：AF=CE；
（2）如果AC=EF，则四边形AFCE是矩形．
27．如图，DB∥AC，且DB=AC，E是AC的中点，
（1）求证：BC=DE；
（2）连接AD、BE，探究：当△ABC满足什么条件时，四边形DBEA是矩形？并说明理由．
28．如图，O是菱形ABCD对角线的交点，作DE∥AC，CE∥BD，DE、CE交于点E，四边形OCED 是矩形吗？说说你的理由．
29．已知：如图，BC是等腰△BED底边ED上的高，四边形ABEC是平行四边形．
求证：四边形ABCD是矩形．
30．如图，已知AB=AC，AD=AE，DE=BC，且∠BAD=∠CAE．
求证：四边形BCED为矩形．
矩形的判定专项练习30题参考答案：
1．（1）∵AD∥BC，
∴∠A+∠B=180°，
∵△DAF≌△CBE，
∴∠A=∠B，
∴2∠A=180°，
∴∠A=90°；
（2）∵AD∥BC，AD=BC，
∴四边形ABCD为平行四边形，
又∵∠A=90°，
∴四边形ABCD是矩形
2．（1）∵∠ABC+∠BCD=180°，BE、CF平分∠ABC，∠BCD，
∴∠GBC+∠GCB=90°，∴∠BGC=90°；（2）∵点H为BC的中点，∴BH=CH=GH，
∵GB∥CM，∴∠BGH=∠CMH，
∵∠HBG=∠HGB，∴∠HCM=∠HMC，
∴MH=BH=CH=GH，
∴四边形GBMC为矩形
3．（1）四边形OCDE是矩形．
证明：∵DE∥AC，CE∥BD，
∴四边形OCED是平行四边形，
又∵AC⊥BD，
∴∠DOC=90°，∴四边形OCED是矩形．
（2）任意改变四边形ABCD的形状，四边形OCED都是平行四边形（答案不唯一）．
理由如下：∵DE∥AC，CE∥BD，
∴四边形OCED是平行四边形．
4．满足△ABC是等腰直角三角形，
∠BAC=90°．
∵△ABC是等腰直角三角形，∠BAC=90°，AD⊥BC于D，
∴BD=CD，
∵点E、F分别是△ABC中AB、AC中点，
∴DF∥AB，ED∥AC，
∴四边形AEDF是平行四边形，
∵∠BAC=90°
∴AEDF是矩形．
5．（1）所作图形如图所示：
（2）四边形DOCE是矩形．
∵△DCE是由△AOB平移后的图形，
∴DE∥AC，CE∥BD．
∴四边形DOCE是平行四边形．
又∵四边形ABCD是菱形，
∴AC⊥BD．即∠DOC=90°
∴四边形DOCE为矩形．
6．∵四边形ABCD为平行四边形，
∴AO=OC，OD=OB，
∵AN=CM ON=OB，
∴ON=OM=OD=OB，
∴四边形NDMB为平行四边形，
∵MN=BD，
∴平行四边形NDMB为矩形
7．∵DE∥AC，CE∥BD，
∴DE∥OC，CE∥OD
∴四边形OCED是平行四边形，
又∵四边形ABCD是菱形，
∴AC⊥BD，
∴∠COD=90°，
∴四边形OCED是矩形
8．（1）证明：∵梯形ABCD中，AD∥BC，即DM∥BE，
∵E、F分别是边BC、CD的中点
∴EF∥BD，
∴四边形DBEM是平行四边形．
（2）证明：连接DE，
∵DB=DC，且E是BC中点，∴DE⊥BC，∴DE∥AB．
又∵AB⊥BC，
∴AB∥DE
∵由（1）知四边形DBEM是平行四边形，∴DM∥BE且DM=BE，
∴DM∥EC且DM=EC，
∴四边形DMCE是平行四边形，
∴CM∥DE，
∴AB∥CM．
又AM∥BC∴四边形ABCM是平行四边形，∵AB⊥BC，∴四边形ABCM是矩形．
9．∵CE平分∠ACB，
∴∠ACE=∠BCE，
∵MN∥BC，
∴∠OEC=∠ECB，
∴∠OEC=∠OCE，
∴OE=OC，
同理，OC=OF，
∴OE=OF．
∵AO=CO，EO=FO，
∴四边形AECF为平行四边形，
∵CE平分∠ACB，
∴∠ACE=∠ACB，
同理，∠ACF=∠ACP，
∴∠ECF=∠ACE+∠ACF=（∠ACB+∠ACP）
=×180°=90°，
∴四边形AECF是矩形．
10．（1）∵BC=2AD，点E是BC的中点，∴EC=AD．
∵AD∥BC，
∴∠ADO=∠CEO，∠DAO=∠ECO．
在△AOD和△COE 中，
∴△AOD≌△COE（ASA）；
（2）∵AD=BE，AD∥BE，
∴四边形ABED是平行四边形；
同理可得：四边形AECD是平行四边形．∴∠ADO=∠B．
∵∠B=∠AOE，
∴∠AOE=2∠B．
∴∠AOE=2∠ADO．
∵∠AOE=∠ADO+∠DAO，
∴∠OAD=∠ODA．
∴OA=OD．
∴AC=DE．
∴四边形AECD是矩形．11．：∵△ABD和△FBC都是等边三角形，∴∠DBF+∠FBA=∠ABC+∠ABF=60°，
∴∠DBF=∠ABC．
又∵BD=BA，BF=BC，
∴△ABC≌△DBF，
∴AC=DF=AE，
同理可证△ABC≌△EFC，
∴AB=EF=AD，
∴四边形DAFEF是平行四边形（两组对边分别相等的四边形是平行四边形）
∵∠BAC=150°，
∴∠DAE=150°﹣∠DAB﹣∠EAC=90°，
∴四边形AEFD为矩形．
12．1）解：∵ABC中AB=BC，∠ABC=90°，BD⊥AC，
∴∠A=∠C=45°，CD=AD，
∴BD=CD=AD，BD平分∠ABC，
∴∠EBD=45°=∠C，
∵BD⊥AC，DE⊥DF，
∴∠BDC=∠EDF=90°，
∴∠BDC﹣∠BDF=∠EDF﹣∠BDF，
∴∠EDB=∠FDC，
∵在△EDB和△FDC中
∴△EDB≌△FDC（ASA），
∴FC=DE=3，
同理△AED≌△BFD，
∴DF=AE=4，
在Rt△EDF中，由勾股定理得：
EF==5；
（2）①证明：∵四边形ABCD是平行四边形，∴AB∥CD，AB=CD，
∵CD=CE，
∴AB∥CE，AB=CE，
∴四边形ABEC是平行四边形，
∴AF=FE，BF=FC，
∵在△ABF和△ECF中
∴△ABF≌△ECF（SSS）；
②证明：∵四边形ABCD是平行四边形，
∴∠ABC=∠D，
∵∠AFC=2∠D，
∴∠AFC=2∠ABC，∵∠AFC=∠ABC+∠FAB，
∵∠ABC=∠FAB，
∴AF=FB，
∵四边形ABCD是平行四边形，
∴AE=2AF，BC=2BF，
∴AE=BC，
∵四边形ABEC是平行四边形，
∴四边形ABEC是矩形．
13．（1）∵AE∥BC，BE∥AD，
∴四边形ADBE是平行四边形，
∴AE=BD，
∵AD是△ABC的中线，
∴BD=CD，
∴AE=CD．
（2）当AB=AC时，四边形ADBE是矩形，理由是：
∵AB=AC，BD=CD，
∴AD⊥BC，即∠ADB=90°，
又∵四边形ADBE是平行四边形，
∴四边形ADBE是矩形
14．1）证明：如图，连接EF．
∵四边形ABCD是梯形，AD∥BC，E、F分别是AB、CD的中点，
∴，EF∥AD∥BC．
∵，
∴EF=CG．
∴四边形EGCF是平行四边形．
∴EG=FC且EG∥FC．
∵F是CD的中点，
∴FC=DF．
∴EG=DF且EG∥DF．
∴四边形DEGF是平行四边形．
（2）证明：连接EF，将EF与DG的交点记为点O．
∵∠ADG=2∠ADE，
∴∠ADE=∠EDG．
∵EF∥AD，
∴∠ADE=∠DEO．
∴∠EDG=∠DEO．
∴EO=DO．
∵四边形DEGF是平行四边形，
∴，．
∴EF=DG，∴平行四边形DEGF是矩形．即四边形DEGF 是矩形．
15．∵点D是AC的中点，
∴DA=DC，
∵AE∥BC，
∴∠AED=∠CFD，
在△ADE和△CDF 中，，
∴△ADE≌△CDF（AAS），
∴AE=CF，
又∵AE∥BC，
∴四边形AECF是平行四边形，
∵AE∥BC，EF∥AB，
∴四边形ABFE是平行四边形，
∴AB=EF，
∵AB=AC，
∴AC=EF，
∴四边形AECF是矩形．
16．∵D、E、F分别是AC、AB、BC的中点，∴DE∥BC，且DE=BC，DF=AB，CF=BC，
∴DE=CF，
∴四边形CFED平行四边形，
又∵CE=AB，
∴CE=DF，
∴平行四边形CFED是矩形，
故四边形CFED是矩形．
17．（1）证明：∵四边形ABCD是平行四边形，
∴AD∥BC，
∴△AEO∽△CFO，
∴=，
∵OA=CO，
∴OE=OF，
∴四边形AECF是平行四边形；
（2）证明：∵四边形AECF是平行四边形，又∵EF⊥AC，
∴平行四边形AECF是菱形；
（3）解：当EF=AC时，四边形AECF是矩形，理由是：由（1）知：四边形AECF是平行四边形，
∵AC=EF，
∴平行四边形AECF是矩形
18．（1）∵DE⊥AC，DF⊥AB，
∴∠AFD=∠AED=∠A=90°，∴四边形AEDF是矩形；
（2）当D时BC的中点时，四边形AEDF是正方形；JU
理由：∵D是BC的中点，
∴BD=DC
∵AB=AC
∴∠B=∠C
又∵DF⊥AB，DE⊥AC，
∴∠BDF=∠DEC
∴△BFD≌△DCE，
∴DF=DE，
∴矩形AEDF是正方形．
19．（1）∵CE平分∠ACB，CF平分∠ACG，∴∠ACE=∠ECB，∠ACF=∠FCG，
又∵MN∥BG，
∴∠DEC=∠ECB，∠DFC=∠FCG，
∴∠DEC=∠DCE，∠DFC=∠DCF，
∴DE=DC，DF=DC，
∴DE=DF．
（2）∵D为AC的中点，
∴AD=DC，
又DE=DF，
∴四边形AECF为平行四边形，
∵∠ACE=∠ECB，∠ACF=∠FCG，
∴∠ECF=90°，
∴平行四边形AECF为矩形
20．∵BE∥AC，CE∥DB，
∴四边形OBEC是平行四边形，
又∵四边形ABCD是菱形，
∴AC⊥BD，
∴∠AOB=90°，
∴平行四边形OBEC是矩形
21．（1）解：OE=OE，
理由是：∵直线l∥BC，
∴∠OEC=∠ECB，
∵CE平分∠ACB，
∴∠OCE=∠BCE，
∴∠OEC=∠OCE，
∴OE=OC，
同理OF=OC，
∴OE=OF．
（2）解：O在AC的中点上时，四边形AECF 是矩形，
理由是：∵OA=OC，OE=OF，
∴四边形AECF是平行四边形，
∵OE=OF=OC=OA，
∴AC=EF，∴平行四边形AECF是矩形
22．（1）证明：∵AF∥BC，
∴∠AFE=∠DCE（1分）
∵E是AD的中点，
∴AE=DE．（2分）
∵∠AEF=∠DEC，
∴△AEF≌△DEC．（3分）
∴AF=DC，
∵AF=BD
∴BD=CD，
∴D是BC的中点；（4分）
（2）四边形AFBD是矩形，（5分）
证明：∵AB=AC，D是BC的中点，
∴AD⊥BC，
∴∠ADB=90°，（6分）
∵AF=BD，AF∥BC，
∴四边形AFBD是平行四边形，（7分）∴四边形AFBD是矩形．
23．∵∠OBC=∠OCB，
∴OB=OC，
∵四边形ABCD是平行四边形，
∴OC=OA=AC，OB=OD=BD，
∴AC=BD，
∵四边形ABCD是平行四边形，
∴四边形ABCD是矩形，
即四边形ABCD是矩形
24．∵ABCD为平行四边形，
∴AD平行且等于BC，
又∵M为AD的中点，N为BC的中点，∴MD平行且等于BN，
∴BNDM为平行四边形，
∴BM∥ND，
同理AN∥MC，
∴四边形PMQN为平行四边形，（5分）连接MN，
∵AM平行且等于BN，
∴四边形ABNM为平行四边形，
又∵AD=2AB，M为AD中点，
∴BN=AB，
∴四边形ABNM为菱形，
∴AN⊥BM，
∴平行四边形PMQN为矩形．（10分）
25．∵四边形ABCD为平行四边形，
∴OA=OC，AE∥FC，
∴∠EAO=∠FCO，
在△AOE和△COF中，
，
∴△AOE≌△COF，
∴AE=CF，
∴四边形AECF为平行四边形，
又∵AF⊥BC，
∴∠AFC=90°，
则四边形AECF为矩形．
26．（1）证明：∵AF∥BE，
∴∠AFD=∠CED，∠FAD=∠DCE，
∵D是AC的中点，
∴AD=DC，
在△FAD和△ECD中
，
∴△FAD≌△ECD（AAS），
∴AF=CE；
（2）证明：∵△FAD≌△ECD，
∴FD=DE，
∵AD=DC，
∴四边形AFCE是平行四边形，
∵AC=EF，
∴平行四边形AFCE是矩形27．（1）证明：∵E是AC的中点，∴EC=AC，
∵DB=AC，
∴DB=EC，
又∵DB∥AC，
∴四边形BCED是平行四边形（一组对边平行且相等的四边形是平行四边形），
∴BC=DE；
（2）解：△ABC满足AB=BC时，四边形DBEA 是矩形．
理由如下：∵E是AC的中点，
∴AE=AC，
∵DB=AC，
∴DB=AE，
又∵DB∥AC，
∴四边形DBEA是平行四边形（一组对边平行且相等的四边形是平行四边形），
∵AB=BC，E为AC中点，
∴∠AEB=90°，
∴平行四边形DBEA是矩形，
即△ABC满足AB=BC时，四边形DBEA是矩形．
28．是矩形．（1分）理由：∵DE∥AC，CE∥BD，
∴四边形OCED是平行四边形，
又∵四边形ABCD是菱形，
∴AC⊥BD，
∴DE⊥CE，
∴∠E=90°，
∴平行四边形OCED是矩形
29．∵BC是等腰△BED底边ED上的高，∴EC=CD，
∵四边形ABEC是平行四边形，
∴AB∥CD，AB=CE=CD，AC=BE，
∴四边形ABCD是平行四边形．
∵AC=BE，BE=BD，
∴AC=BD，
∴四边形ABCD是矩形
30．在△ABD和△ACE中，
∵AB=AC，AD=AE，∠BAD=∠CAE，
∴△ABD≌△ACE（SAS）
∴BD=CE又DE=BC．
∴四边形BCED为平行四边形．在△ACD和△ABE中，
∵AC=AB，AD=AE，
∠CAD=∠CAB+∠BAD=∠CAB+∠CAE=∠BAE，
∴△ADC≌△AEB（SAS），∴CD=BE．∴四边形BCED为矩形．（对角线相等的平行
四边形是矩形）
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1. I can’t swim b ut I have a strong fear of water. Look back at my childhood, I think that three reasons might explain fear. Firstly, I am not allowed to go near the water when I was a child, as my mother had an unreasonable fear of it. Therefore, I was taught to see to the water as something danger. Secondly, my eyes became bad when I was five. If I took off my glass in the water, I couldn’t see something, and this increased my fear. What’s worse, for a child I once saw a neighbor drowned. Since then I have been more frighte ning. 2. Boys and girls, May I have your attention, please? I have anything important to tell you. Next Sunday afternoon , our class was going to hold an English story-telling competition in lecture hall. Everybody is required to take part in the activity. Two foreign teacher will be invited to act as judges. The activity will benefit from you in many ways. For example, it will give us a good chance to practicing your oral English and train your communication skills. However, it will inspire your interest at English. The requirements are as follows. First, the story can be original. That is, the story must be made up by yourself. Second, the story must be . presented in English. Finally, your story must be finished within three minutes. That’s all, thank you 3. Dear Jack, I am pleasing to hear from you and know that you had a greatly time working part-time in a fast-food restaurant. Now I would like to share your working experiences with you during the past summer. In the beginning of the summer vacation, I was hired as shopping guide in a big supermarket. I stood for ten hours a day and smiled to everyone I have met. Sometimes I was laughed at, and most of the time, I was praised for my wonderful works by many people. Half a month passed quickly after I knew it, and I learned how to deal with different people. Above all, I have learned that smile is the key to solve any problem. Best wishes, Yours Sincerely, Li Ming
短文改错专项练习高考真题
预习案 近年高考，英语短文改错题，学生和老师都认为难度不大，然而，每年高考中部分同学还是对此深感头疼， 主要原因：一、英语基础知识不扎实 二、对短文改错题的考点不了解 三、最欠缺是行文逻辑能力，学生上下文联系呼应及翻译能力有待增强 目标：1、学生能加强对短文改错题型的了解， 2、通过针对性练习，学生能增强自信心和做题的准确度 重点：时态语态、逻辑一致 难点：逻辑一致 （每题一处错误） 1.My father and I stayed at the South Lake Hotel for a week when we visit Beijing last month. 2.The book you borrowed from the library should returned in five days. 3.Therefore, I told them interested stories and how I was enjoying Brazil. 4.In early January this year, the rate of UFO reports was steady, around three per weeks. 5.Unfortunate, the only problem was the journey home. 6.I knew that they would be worried about me because I was so farther away. 7.I’m sorry that I am abroad and can’t send your flowers. 8.There was one on particular I’d always wanted. 9.It is such great hotel that I would recommend it to any friend of mine. 10.In the end, we drove to a service station and waited there unless the road was clear. 11.He said it was best to stay until help arrived rather than go into the forest and getting lost. 预习案 近年高考，英语短文改错题，学生和老师都认为难度不大，然而，每年高考中部分同学还是对此深感头疼， 主要原因：一、英语基础知识不扎实 二、对短文改错题的考点不了解 三、最欠缺是行文逻辑能力，学生上下文联系呼应及翻译能力有待增强 目标：1、学生能加强对短文改错题型的了解， 2、通过针对性练习，学生能增强自信心和做题的准确度 重点：时态语态、逻辑一致 难点：逻辑一致 （每题一处错误） 1.My father and I stayed at the South Lake Hotel for a week when we visit Beijing last month. 2.The book you borrowed from the library should returned in five days. 3.Therefore, I told them interested stories and how I was enjoying Brazil. 4.In early January this year, the rate of UFO reports was steady, around three per weeks. 5.Unfortunate, the only problem was the journey home. 6.I knew that they would be worried about me because I was so farther away. 7.I’m sorry that I am abroad and can’t send your flowers. 8.There was one on particular I’d always wanted. 9.It is such great hotel that I would recommend it to any friend of mine. 10.In the end, we drove to a service station and waited there unless the road was clear. 11.He said it was best to stay until help arrived rather than go into the forest and getting lost. 探究案