1.如图1、2是两个相似比为1 :2的等腰直角三角形,将两个三角形如图3放置,小直角三角形的斜边与大直角三角形的一直角边重合. (1)在图3中,绕点D 旋转小直角三角形,使两直角边分别与AC 、BC 交于点E 、F ,如图4.
求证:AE 2
+BF 2
=EF 2
;
(2)若在图3中,绕点C 旋转小直角三角形,使它的斜边和CD 延长线分别与AB 交于点E 、F ,如图5,此时结论AE 2+BF 2=EF 2
是否仍然成立?若成立,请给出证明;若不成立,请说明理由;
(3)如图,在正方形ABCD 中,E 、F 分别是边BC 、CD 上的点,满足△CEF 的周长等于正方形ABCD 的周长的一半,AE 、AF 分别与对角线BD 交于M 、N ,试问线段BM 、MN 、DN 能否构成三角形的三边长?若能,指出三角形的形状,并给出证明;若不能,请说明理由. ;
B 图2 A
C B 图3 A C
D 图1 B 图4 A C D E
F B 图5 A C
D
E F A D
F
M
N
解:(1)如图4,由于AD =BD ,将△AED 绕点D 旋转180°,得△BE ′D
则AE =BE ′,ED =E ′D ,连接E ′F
∵∠FBE ′=∠ABC +∠ABE ′=∠ABC +∠CAB =90°
∴在Rt △BE ′F 中有BE ′ 2+BF 2=E ′F 2
又∵FD 垂直平分EE ′,∴EF =E ′F
∴AE 2
+BF 2
=EF 2
························································ 6分
(2)如图5,由于AC =BC ,将△AEC 绕点C 旋转90°,得△BE ′C 则AE =BE ′,CE =CE ′,连接E ′F
∵∠FBE ′=∠ABC +∠CBE ′=∠ABC +∠CAB =90°
∴在Rt △BE ′F 中有BE ′ 2+BF 2=E ′F 2
∵∠E ′CF =∠E ′CB +∠BCF =∠ACE +∠BCF
=90°-∠ECF =90°-45°=45°=∠ECF
CE =CE ′,CF =CF
∴△CEF ≌△CE ′F ,∴EF =E ′F
∴AE 2
+BF 2
=EF 2
······································································································ 12分 (3)将△ADF 绕点A 顺时针旋转90°,得△ABG ,且FD =GB ,AF =AG
B
图5
A
C
D
E
F
E ′
因为△CEF 的周长等于正方形ABCD 的周长的一半,所以 CE +EF +CF =CD +CB =CF +FD +CE +BE ∴EF =FD +BE =GB +BE =GE
从而可得△AEG ≌△AEF ,∴∠EAG =∠EAF 又∵∠EAG =∠EAB +∠BAG ,∠BAG =∠DAF
∴∠EAF =∠EAB +∠DAF ,而∠EAB +∠EAF +∠DAF =90° ∴∠EAF =45°
由(2)知BM 2
+DN 2
=MN
2
∴由勾股定理的逆定理知:线段BM 、MN 、DN 能构成直角三角形 ······················· 18分 2.(1)操作发现·
如图,矩形ABCD 中,E 是AD 的中点,将△ABE 沿BE 折叠后得到△GBE ,且点G 在矩形ABCD 内部.小明将BG 延长交DC 于点F ,认为GF =DF ,你同意吗?说明理由.
(2)问题解决 保持(1)中的条件不变,若DC =2DF ,求AB
AD
的值; (3)类比探究
保持(1)中的条件不变,若DC =n ·DF ,求
AB
AD
的值. 解:(1)同意.连接EF ,则∠EGF =∠D =90°,EG =AE =ED ,EF =EF
∴Rt △EGF ≌Rt △EDF ,∴GF =DF ············································································ 3分 (2)由(1)知GF =DF ,设DF =x ,BC =y ,则有GF =x ,AD =y ∵DC =2DF ,∴CF =x ,DC =AB =BG =2x
A
C
D
F
M
N
∴BF =BG +GF =3x
在Rt △BCF 中,BC 2
+CF 2
=BF 2
,即y 2
+x 2
=(3x )2
∴y =22x ,∴
AB AD =x
y 2=2 ············································· 6分 (3)由(1)知GF =DF ,设DF =x ,BC =y ,则有GF =x ,AD =y ∵DC =n ·DF ,∴DC =AB =BG =nx ∴CF =(n -1)x ,BF =BG +GF =(n +1)x
在Rt △BCF 中,BC 2
+CF 2
=BF 2
,即y 2
+[(n -1)x ]2
=[(n +1)x ]2
∴y =n 2x ,∴
AB AD
=nx y =n n 2(或n
2) ·························································· 10分
3.如图1所示,在直角梯形ABCD 中,AD ∥BC ,AB ⊥BC ,∠DCB =75o,以CD 为一边的等边△DCE 的另一顶点E 在腰AB 上.
(1)求∠AED 的度数; (2)求证:AB =BC ;
(3)如图2所示,若F 为线段CD 上一点,∠FBC =30o.
求
DF
FC
的值.
解:(1)∵∠BCD =75o,AD ∥BC ,∴∠ADC =105o ·························································· 1分
A
B C D
E 图1 A
B C D
E 图2
F
由等边△DCE可知:∠CDE=60o,故∠ADE=45o
由AB⊥BC,AD∥BC可得:∠DAB=90o,∴∠AED=45o ·························· 3分(2)方法一:由(1)知:∠AED=45o,∴AD=AE,故点A在线段DE的垂直平分线上∴AC就是线段DE的垂直平分线,即AC⊥DE ··········································· 5分
连接AC,∵∠AED=45o,∴∠BAC=45o
又∵AB⊥BC,∴AB=BC ··················································7分
方法二:过D点作DH⊥BC于H,则DH=AB ················4分
易证△DHC≌△CBE,∴DH=BC····································6分
∴AB=BC···········································································7分
(3)∵∠FBC=30o,∴∠ABF=60o
连接AF,设BF、AD的延长线相交于点G
∵∠FBC=30o,∠DCB=75o,∴∠BFC=75o,故BC=BF
由(2)知:BA=BC,故BA=BF
∵∠ABF=60o,∴AB=BF=F A,
又∵AD∥BC,AB⊥BC,∴∠F AG=∠G=30o
∴FG=F A=FB ············································ 10分
∵∠G=∠FBC=30o,∠DFG=∠CFB,FB=FG
∴△BCF≌△GDF ········································11分
∴DF=CF,即点F是线段CD的中点
∴DF
FC=1 ········································································································· 12分
A
B C
D
E
图1
H
A
B C
D
E
F
G
4.如图①,在等腰梯形ABCD 中,AD ∥BC ,AE ⊥BC 于点E ,DF ⊥BC 于点F .AD =2cm ,BC =6cm ,AE =4cm .点P 、Q 分别在线段AE 、DF 上,顺次连接B 、P 、Q 、C ,线段BP 、PQ 、QC 、CB 所围成的封闭图形记为M .若点P 在线段AE 上运动时,点Q 也随之在线段DF 上运动,使图形M 的形状发
生改变,但面积始终..为10cm 2
.设EP =x cm ,FQ =y cm ,解答下列问题: (1)直接写出当x =3时y 的值;
(2)求y 与x 之间的函数关系式,并写出自变量x 的取值范围; (3)当x 取何值时,图形M 成为等腰梯形?图形M 成为三角形? (4)直接写出线段PQ 在运动过程中所能扫过的区域的面积.
解:(1)y =2 ·························································································································· 2分
(2)由等腰梯形的性质得BE =EF =FC =2
∵S △BEP +S 梯形PEFQ +S △FCQ =S 图形M
∴
21×2x +21(
x +y )×2+21
×2y =10 ∴y =-x +5 ···································································································· 4分 由?
????0≤x
≤40≤-x +5
≤4 得1≤x
≤4 ····································································· 5分 (3)若图形M 为等腰梯形(如图①),则EP =FQ
A B C
D
E
F
(备用图)
A B C D E F Q
P
图①
即x =-x +5,解得x =2
5 ∴当x =
2
5
时,图形M 为等腰梯形 ······························································ 6分 若图形M 为三角形,分两种情形:
①当点P 、Q 、C 在一条直线上时(如图②),EP 是△BPC 的高 ∴21BC ·EP =10,即21×6x =10,解得x =3
10
············································· 7分
②当点B 、P 、Q 在一条直线上时(如图③),FQ 是△BQC 的高 ∴
21BC ·FQ =10,即21×6×(
-x +5
)=10,解得x =3
5 ····························· 8分 ∴当x =310或3
5
时,图形M 为三角形
A
B
C D
E
F
Q
P
图③
A B
C
D E
F Q
P
图①
A B
C
D
E
F
Q
P
图②
C
图④
(4)3cm 2 ·············································································································· 10分
评分说明:(4)中不写单位不扣分,线段PQ 在运动过程中所能扫过的区域为图④中的阴影部分
5.(贵州省铜仁地区)已知:如图,在Rt △ABC 中,∠ACB =90°,∠A =30°,CD ⊥AB 交AB 于点E ,且CD =AC ,DF ∥BC 分别与AB 、AC 交于点G 、F .
(1)求证:GE =GF ; (2)若BD =1,求DF 的长.
解:(1)证明:∵DF ∥BC ,∠ACB =90°,∴∠CFD =
∵CD ⊥AB ,∴∠AEC =90° 在Rt △AEC 和Rt △DFC 中,
∵∠AEC =∠CFD =90°,∠ACE =∠DCF ,DC =AC ∴Rt △AEC ≌Rt △DFC ·········································· 3分 ∴CE =CF
在Rt △AEC 中,∵∠A =30°,∴CE =21AC =2
1
DC ∴DE =AF
又∠DGE =∠AGF ,∠DEG =∠AFG =90° ∴Rt △DEG ≌Rt △AFG
D
∴GE =GF ·························································· 6分
(2)解:∵CD ⊥AB ,CE =ED ,∴BC =BD
又∠EDB =∠ECB =∠A =30°,∠DEB =90°,BD =1 ∴DE =
23BD =2
3
···················································································· 8分 ∴CD =2DE =3 ························································································ 10分 ∵DF ∥BC ,∴∠CDF =∠ECB =30°
∴DF =
23CD =2
3
6.(山东省威海市)如图①,将一张矩形纸片对折,然后沿虚线剪切,得到两个(不等边)三角形纸片△ABC ,△A 1B 1C 1.
(1)将△ABC ,△A 1B 1C 1如图②摆放,使点A 1与B 重合,点B 1在AC 边的延长线上,连接CC 1交BB 1于点E .求证:∠B 1C 1C =∠B 1BC .
图 ①
A B
C A 1
B 1
C 1
C 1 1 B (A 1) E
图 ②
(2)若将△ABC ,△A 1B 1C 1如图③摆放,使点B 1与B 重合,点A 1在AC 边的延长线上,连接CC 1交A 1B 于点F .试判断∠A 1C 1C 与∠A 1BC 是否相等,并说明理由.
解:(1)证明:由题意,知△ABC ≌△A 1B 1C 1
∴AB =A 1B 1,BC 1=AC ,∠2=∠7,∠A =∠1 ∴∠3=∠A =∠1 ····································· 1分 ∴ C 1∥AC
∴四边形ABC 1C 是平行四边形 ··············· 2分 ∴AB ∥CC 1
∴∠4=∠7=∠2 ····················································································· 3分 ∵∠5=∠6
∴∠B 1C 1C =∠B 1BC ················································································ 4分 (2)∠A 1C 1C =∠A 1BC ···················································································· 5分
理由如下:由题意,知△ABC ≌△A 1B 1C 1 ∴AB =A 1B 1,BC 1=BC ,∠1=∠8,∠A =∠2 ∴∠3=∠A ,∠4=∠7 ···························· 6分
C 1
1 B (A 1)
E
图 ②
1
2
3
4
5
6
7
C 1
C A 1 B (B 1)
F
图 ③
1 8
7 5
6
4
2 3
C 1 1 B (B 1) F 图 ③
∴∠1+∠FBC =∠8+∠FBC
∴∠C 1BC =∠A 1BA ·································· 7分
∵∠4=21(180°-∠C 1BC ),∠A =2
1
(180°-∠A 1BA )
∴∠4=∠A ······························································································· 8分 ∴∠4=∠2 ∵∠5=∠6
∴∠A 1C 1C =∠A 1BC ················································································ 8分
7.(山东省淄博市)将一副三角尺如图拼接:含30°角的三角尺(△ABC )的长直角边与含45°角的三角尺(△ACD )的斜边恰好重合.已知AB =32,P 是AC 上的一个动点.
(1)当点P 运动到∠ABC 的平分线上时,连接DP ,求DP 的长; (2)当点P 在运动过程中出现PD =BC 时,求此时∠PDA 的度数;
(3)当点P 运动到什么位置时,以D ,P ,B ,Q 为顶点的平行四边形的顶点Q 恰好在边BC 上?求出此时□DPBQ 的面积.
解:(1)如图(1),作DF ⊥AC 于F
在Rt △ABC 中,∵AB =32,∠BAC =30°,∴BC =3,AC =3
在Rt △ACD 中,∵AD =CD ,∴DF =AF =CF =23
∵BP 平分∠ABC ,∴∠PBC =30°
A B
C D
∴CP =BC ·tan30°=1,∴PF =2
1
∴DP =22PF DF
+=
2
10 ······································································ 3分
(2)当P 点位置如图(2)所示时,根据(1)中结论,DF =
2
3
,∠ADF =45° 又PD =BC =3,∴cos ∠PDF =
PD
DF =23,∴∠PDF =30°
∴∠PDA =∠ADF -∠PDF =15° ································································ 5分 当P 点位置如图(3)所示时,同(2)可得∠PDF =30°
∴∠PDA =∠ADF +∠PDF =75° ································································ 7分
(2)
(1)
(3)
(4)
(3)当CP =
2
3
时,以D ,P ,B ,Q 为顶点的平行四边形的顶点Q 恰好在边BC 上 理由如下:
如图(4),在□DPBQ 中,∵BC ∥DP ,∠ACB =90°,∴DP ⊥AC 根据(1)中结论可知,DP =CP =2
3
························································ 8分 ∴S □DPBQ
=DP ·CP =
4
9 ············································································· 10分 8.如图1,已知矩形ABED ,点C 是边DE 的中点,且AB =2AD . (1)判断△ABC 的形状,并说明理由;
(2)保持图1中ABC 的固定不变,绕点C 旋转DE 所在的直线MN 到图2中的位置(当垂线段AD 、BE 在直线MN 的同侧),试探究线段AD 、BE 、DE 长度之间有什么关系?并给予证明;
(3)保持图2中△ABC 的固定不变,继续绕点C 旋转DE 所在的直线MN 到图3中的位置(当垂线段AD 、BE 在直线MN 的异侧).试探究线段AD 、BE 、DE 长度之间有什么关系?并给予证明.
A
B
C D E 图1
A
B
C
D E 图3
N M A
B
C D
E 图2
N
M
解:(1)△ABC 为等腰直角三角形 ····················································································· 1分
如图1,在矩形ABED 中
∵点C 是边DE 的中点,且AB =2AD ∴AD =DC =CE =EB ,∠D =∠E =90?
∴Rt △ADC ≌Rt △BEC ····································· 2分 ∴AC =BC ,∠1=∠2=45? ∴∠ACB =90?
∴△ABC 为等腰直角三角形 ········································································ 3分 (2)DE =AD +BE ································································································ 4分
如图2,在Rt △ADC 和Rt △CEB 中 ∵∠1+∠CAD =90?,∠1+∠2=90? ∴∠CAD =∠2
又∵AC =CB ,∠ADC =∠CEB =90?
∴Rt △ADC ≌Rt △CEB ····································· 6分 ∴DC =BE ,CE =AD
∴DC +CE =BE +AD ,即DE =AD +BE ····················································· 7分 (3)DE =BE -AD ································································································ 8分
如图3,Rt △ADC 和Rt △CEB 中 ∵∠1+∠CAD =90?,∠1+∠2=90?
A
B
C D
E
图1
1
2
A
B
C
D
E 图2 N
M 1 2
A
B C
D E
图3
N
M 1 2
∴∠CAD =∠2
又∵∠ADC =∠CEB =90?,AC =CB
∴Rt △ADC ≌Rt △CEB ··································· 10分 ∴DC =BE ,CE =AD ∴DC -CE =BE -AD
即DE =BE -AD ························································································· 11分
9.如图1,梯形ABCD 中,AD ∥BC ,AB =AD =CD ,∠BAD =120°,∠MAN =60°,将图1中的∠MAN 绕点A 按逆时针方向旋转α角(0°<α<120°),边AM 、AN 分别交直线BC 、CD 于E 、F 两点. (1)当0°<α≤60°时,其他条件不变,如图2、如图3所示.
①如图2,判断线段BE 、DF 、EF 的数量关系,并直接写出结论;
②如图3,①中的结论是否依然成立?若成立,请利用图3证明;若不成立,说明理由. (2)当60°<α<120°时,其他条件不变,请在图4中画出一个..符合条件的图形,直接写出所画图形中线段BE 、DF 、EF 的数量关系.
图1 A
B C D
M
N 图2 A
B
C D
M
N
E F
图3
A
B
C
D M F
E
N
图4
A
B
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解:(1)①BE +DF =EF ······································································································ 4分
②①中的结论依然成立 ················································································ 6分 证明:延长CD 到点G ,使得DG =BE ,连接AG 、EG ,交AN 于点H ,如图3 ∵∠B =∠ADG =60°,AB =AD ,BE =DG ∴ABE △≌ADG ,∴AE =AG ,∠BAE =∠DAG 又∵∠BAE +∠F AD =60°,∴∠DAG +∠F AD =60° 即∠F AG =∠MAN =60° ∴AH 垂直且平分EG ∴GF =EF ,即GD +DF =EF
∴BE +DF =EF ····························································································· 8分 (2)如图4,BE -DF =EF ······································································· 12分
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10.如图,△ABC 是等边三角形,点D 是边BC 上的一点,以AD 为边作等边△ADE ,过点C 作CF ∥DE 交AB 于点F . (1)若点D 是BC 边的中点(如图①),求证:EF =CD ; (2)在(1)的条件下直接写出△AEF 和△ABC 的面积比; (3)若点D 是BC 边的任意一点(除B 、C 外如图②),那么,(1)中的结论是否仍然成立?若成立,请给出证明;若不成立,请说明理由.
解:(1)证明:∵△ABC 是等边三角形,D 是BC 边的中点
∴AD ⊥BC ,且∠DAB =2
1
∠BAC =30°
∵△AED 是等边三角形 ∴AD =AE ,∠ADE =60° ∴∠EDB =30°
∵ED ∥CF ,∴∠FCB =∠EDB =30°
∵∠ACB =60°,∴∠ACF =∠BAD =30° ························································ 2分 又∵∠B =∠F AC =60°,AB =CA
∴△ABD ≌△CAF ································································································ 3分 ∴AD =CF
∵AD =ED ,∴ED =CF ,又∵ED ∥CF
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D 图① F
E A B C D 图②
∴四边形EDCF 是平行四边形
∴EF =CD ············································································································ 4分 (2)S △AEF :
S △ABC
=1 :
4 ······························································································ 6分
(3)成立 ······················································································································ 7分
证明:∵ED ∥FC ,∴∠EDB =∠FCB ∴∠ACF =∠ACB -∠FCB =60°-∠FCB
△ABD 中,∠BAD =180°-∠B -∠EDB -∠ADE =60°-∠EDB ∴∠ACF =∠BAD ················································ 9分 又∵∠B =∠F AC =60°,AB =CA
∴△ABD ≌△CAF ·············································· 10分 ∴AD =FC
∵AD =ED ,∴ED =CF ,又∵ED ∥CF ∴四边形EDCF 是平行四边形
∴EF =DC ·········································································································· 12分
11.如图1,Rt △ABC 中,∠ABC =90°,BC <AB <2BC .在AB 边上取一点M ,使AM =BC ,过点A 作AD ⊥AB 且AD =BM ,连接DC ,再过点A 作AN ∥DC ,交直线CM 、CB 于点E 、N . (1)求证:∠AEM =45°;
(2)若将题中的条件“BC <AB <2BC ”改为“AB >2BC ”,其它条件不变,请在图2中画出图形,此时(1)中的结论是否仍然成立,若成立,请说明理由;若不成立,请猜想∠AEM 的度数,并说明理由.
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解:(1)证明:连接DM ,如图1
∵AD ⊥AB ,∴∠DAM =∠B =90° ∵AD =BM ,AM =BC ,∴△ADM ≌△BMC ∴∠ADM =∠BMC ,DM =MC
∵∠ADM +∠AMD =90°,∴∠BMC +∠AMD =90° ∴∠DMC =90°
∴△DMC 是等腰直角三角形 ∴∠DCM =45° ∵AN ∥DC
∴∠AEM =∠DCM =45° ····················································································· 5分 (2)画出图形,如图2 ································································································ 7分
不成立.∠AEM =135° ························································································ 8分 连接DM
由(1)知∠DCM =45°
∵AN ∥DC ,∴∠AEM +∠DCM =180°
∴∠AEM =135° ················································· 10分
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12.在△ABC 中,BA =BC ,∠BAC =α,M 是AC 的中点,P 是线段BM 上的动点,将线段P A 绕点P 顺时针旋转2α得到线段PQ . (1)若α=60°且点P 与点M 重合(如图1),线段CQ 的延长线交射线BM 于点D ,请补全图形,并写出∠CDB 的度数; (2)在图2中,点P 不与点B ,M 重合,线段CQ 的延长线与射线BM 交于点D ,猜想∠CDB 的大小(用含α的代数式表示),并加以证明;
(3)对于适当大小的α,当点P 在线段BM 上运动到某一位置(不与点B ,M 重合)时,能使得线段CQ 的延长线与射线BM 交于点D ,且PQ =QD ,请直接写出α的范围.
解:(1)补全图形,见图1;∠CDB =30° (2)猜想:∠CDB =90°-α 证明:如图2 ,连结AD ,PC
∵BA =BC ,M 是AC 的中点,∴BM ⊥AC
∵点D ,P 在直线BM 上,∴P A =PC ,DA =DC 又∵DP 为公共边,∴△ADP ≌△CDP ∴∠DAP =∠DCP ,∠ADP =∠CDP 又∵P A =PQ ,∴PQ =PC
∴∠DCP =∠PQC ,∠DAP =∠PQC
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A B C Q M (P ) 图2
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九年级数学几何模型压轴题专题练习(解析版) 一、初三数学 旋转易错题压轴题(难) 1.已知:如图①,在矩形ABCD 中,3,4,AB AD AE BD ==⊥,垂足是E .点F 是点 E 关于AB 的对称点,连接A F 、BF . (1)求AF 和BE 的长; (2)若将ABF 沿着射线BD 方向平移,设平移的距离为m (平移距离指点B 沿BD 方向所经过的线段长度).当点F 分别平移到线段AB AD 、上时,直接写出相应的m 的值. (3)如图②,将ABF 绕点B 顺时针旋转一个角1(080)a a ?<,记旋转中ABF 为 ''A BF ,在旋转过程中,设''A F 所在的直线与直线AD 交于点P ,与直线BD 交于点Q .是否存在这样的P Q 、两点,使DPQ 为等腰三角形?若存在,求出此时DQ 的 长;若不存在,请说明理由. 【答案】(1)129,55AF BF = =;(2)95 m =或16 5m =;(3)存在4组符合条件的点 P 、点Q ,使DPQ 为等腰三角形; DQ 的长度分别为2或 258 9 1055或3 5105 【解析】 【分析】 (1)利用矩形性质、勾股定理及三角形面积公式求解; (2)依题意画出图形,如图①-1所示.利用平移性质,确定图形中的等腰三角形,分别求出m 的值; (3)在旋转过程中,等腰△DPQ 有4种情形,分别画出图形,对于各种情形分别进行计算即可. 【详解】 (1)∵四边形ABCD 是矩形, ∴∠BAD=90°, 在Rt △ABD 中,AB=3,AD=4, 由勾股定理得:2222345AB AD +=+=, ∵S △ABD 12= BD?AE=1 2 AB?AD ,
1.(1)操作发现· 如图,矩形ABCD 中,E 是AD 的中点,将△ABE 沿BE 折叠后得到△GBE ,且点G 在矩形ABCD 内部.小明将BG 延长交DC 于点F ,认为GF =DF ,你同意吗?说明理由. (2)问题解决 保持(1)中的条件不变,若DC =2DF ,求AB AD 的值; (3)类比探究 保持(1)中的条件不变,若DC =n ·DF ,求 AB AD 的值. 2.如图1所示,在直角梯形ABCD 中,AD ∥BC ,AB ⊥BC ,∠DCB =75o,以CD 为一边的
等边△DCE 的另一顶点E 在腰AB 上. (1)求∠AED 的度数; (2)求证:AB =BC ; (3)如图2所示,若F 为线段CD 上一点,∠FBC =30o. 求 DF FC 的值. 3.如图①,在等腰梯形ABCD 中,AD ∥BC ,AE ⊥BC 于点E ,DF ⊥BC 于点F .AD =2cm ,BC =6cm ,AE =4cm .点P 、Q 分别在线段AE 、DF 上,顺次连接B 、P 、Q 、C ,线段BP 、PQ 、QC 、CB 所围成的封闭图形记为M .若点P 在线段AE 上运动时,点Q 也随之在线段DF 上运动,使图形M 的形状发生改变,但面积始终.. 为10cm 2.设EP =x cm ,FQ =y cm ,A B C D E 图1 A B C D E 图2 F
解答下列问题: (1)直接写出当x =3时y 的值; (2)求y 与x 之间的函数关系式,并写出自变量x 的取值范围; (3)当x 取何值时,图形M 成为等腰梯形?图形M 成为三角形? (4)直接写出线段PQ 在运动过程中所能扫过的区域的面积. 4.如图①,将一张矩形纸片对折,然后沿虚线剪切,得到两个(不等边)三角形纸片△ABC ,△A 1B 1C 1. A B C D E F (备用图) A B C D E F Q P 图① 图 ① A C A 1 B 1 C 1
第8章 第1节 一、选择题 1.(2010·崇文区)“m =-2”是“直线(m +1)x +y -2=0与直线mx +(2m +2)y +1=0相互垂直”的( ) A .充分不必要条件 B .必要不充分条件 C .充要条件 D .既不充分也不必要条件 [答案] A [解析] m =-2时,两直线-x +y -2=0、-2x -2y +1=0相互垂直;两直线相互垂直时,m(m +1)+2m +2=0,∴m =-1或-2,故选A. 2.(文)(2010·安徽文)过点(1,0)且与直线x -2y -2=0平行的直线方程是( ) A .x -2y -1=0 B .x -2y +1=0 C .2x +y -2=0 D .x +2y -1=0 [答案] A [解析] 解法1:所求直线斜率为12,过点(1,0),由点斜式得,y =12(x -1),即x -2y -1=0. 解法2:设所求直线方程为x -2y +b =0, ∵过点(1,0),∴b =-1,故选A. (理)设曲线y =ax2在点(1,a)处的切线与直线2x -y -6=0平行,则a =( ) A .1 B.12 C .-12 D .-1 [答案] A [解析] y′=2ax ,在(1,a)处切线的斜率为k =2a , 因为与直线2x -y -6=0平行,所以2a =2,解得a =1. 3.点(-1,1)关于直线x -y -1=0的对称点是( ) A .(-1,1) B .(1,-1) C .(-2,2) D .(2,-2) [答案] D [解析] 一般解法:设对称点为(x ,y),则