15.(5分)如图,在ABCD中,E为CD的中点,连结AE并延长交BC的延长线于点F.求证:S△ABFSABCD.
15.证明:四边形ABCD为平行四边形,AD∥BC.
∠DAE∠F,∠D∠ECF.
E是DC的中点,DECE.
△AED≌△FEC. ································································································· 3分 S△AEDS△FEC. A
D E S△ABFS四边形ABCES△CEF
S四边形ABCES△AED
SABCD 5分 B C F
21.(9分)如图,AB为O的直径,AC,BD分别和O相切于点A,B,点E为圆上不与A,B重合的点,过点E作O的切线分别交AC,BD于点C,D,连结OC,OD分别交AE,BE于点M,N.
(1)若AC4,BD9,求O的半径及弦AE的长;
(2)当点E在O上运动时,试判定四边形OMEN的形状,并给出证明.
21.解:(1)AC,BD,CD分别切O于A,B,E,AC4,BD9, A C CEAC4,DEBD9.
CD13. M
AB为O的直径,∠BAC∠ABD90.
过点C作CF⊥BD于F,则四边形ABFC是矩形. E O N
B D
FD
5,CF12.
连结OE. AB12,O的半径为6. ················································································ 3分
CACE,OAOE,
OC垂直平分弦AE.
OC
AMAOAC
OC ··························································································· 6分 AE2AM
(2)当点E在O上运动时,由(1)知OC垂直平分AE.同理,OD垂直平分BE. AB为直径,∠AEB90.四边形OMEN为矩形. ··································· 8分 当动点E满足OE⊥AB时,OAOE,∠OEA45.
MOME.
矩形OMEN为正方形
20.(9分)如图,ABCD是边长为1的正方形,其中DE、
⌒
EF、FG的圆心依次是点A、B、C.
(1)求点D沿三条圆弧运动到G所经过的路线长;
(2)判断直线GB与DF的位置关系,并说明理由.
F
⌒⌒
B
A
E
D
20.解:(1)∵AD = 1,∠DAE = 90o,
901
, 180
2的长l902, 同理,EF2
180
的长l9033, FG
31802
的长l∴DE
1所以,点D运动到点G所经过的路线长ll1l2l33.
(2)直线GB⊥DF.
理由如下:延长GB交DF于H.
∵CD = CB,∠DCF = ∠BCG,CF = CG, ∴△FDC≌△GBC. ∴∠F =∠G.
o
又∵∠F + ∠FDC = 90,
o
∴∠G + ∠FDC = 90,
o
即∠GHD =90,故 GB⊥DF.
17.(9分)如图,点E、F、G分别 是□ABCD的边AB、BC、CD、DA的中点.求证:ΔBEF≌ΔDGH.
17.证明:∵四边形ABCD是平行四边形,
A∴∠B = ∠D,AB = CD,BC = AD .
又∵E、F、G、H分别是平行四边形ABCD ∴BE = DG,BF = DH.
B∴△BEF≌△DGH. 21.(10分)请你画出一个以BC为底边的等腰ΔABC,使底边上的高AD=BC.
(1)求tanB和 sinB的值;
(2)在你所画的等腰ΔABC中设底边BC=5米,求腰上的高BE.
21.解:如图,正确画出图形.
(1)∵AB = AC,AD⊥BC,AD = BC,
∴BD∴AB∴tanB
1
1BCAD.即 AD = 2BD.
2
2.
AD
2, BD
A
sinB
AD.
AB(2)作BE⊥AC于E.
在Rt△BEC
中,sinCsinABC.
又∵sinC
D
C
BE
,
BC
BE
.
故BE.
17.(9分)如图所示,∠BAC=∠ABD,AC=BD,点O是AD、BC的交点,点E是AB的中点.试判断OE和AB的位置关系,并给出证明.17.OE⊥AB.„„„„„„„„„„„„„„„„1分 证明:在△BAC和△ABD中,
AC=BD,∠BAC=∠ABD,AB=BA.∴
△
BAC
≌
△
ABD.„„„„„„„„„„„„„„„„„„„„„5分
∴∠OBA=∠OAB,
∴OA=OB.„„„„„„„„„„„„„„„„„„„„„7分 又∵AE=BE, ∴OE⊥AB.„„„„„„„„„„„„„„„„„„„„„9分 (注:若开始未给出判断“OE⊥AB”,但证明过程正确,不扣分)
21.(10分)如图,在Rt△ABC中,∠ACB=90°, ∠B =60°,BC=2.点0是AC的中点,过
点0的直线l从与AC重合的位置开始,绕点0作逆时针旋转,交AB边于点D.过
点C作CE∥AB交直线l于点E,设直线l的旋转角为α.
(1)①当α=________度时,四边形EDBC是等腰梯形,此时AD的长为_________;②当α=________度时,四边形EDBC是直角梯形,此时AD的长为_________;(2)当α=90°时,判断四边形EDBC是否为菱形,并说明理由. 21.
(
)
①
30
,
;
②
60
,
1.5;„„„„„„„„4分(2)当∠α=90时,四边形EDBC是菱形.∵∠α=∠ACB=90,∴BC//ED.∵CE//AB, ∴四边形形.„„„„„„„„6分
在Rt△ABC中,∠ACB=90,∠B=60,BC=2,
∴∠A=30.
EDBC是平行四边
∴AB=4,AC
∴AO=
AC
„„„„„„„„8分
2在Rt△AOD中,∠A=30,∴AD=2.∴BD=2.∴BD=BC.
又∵四边形EDBC是平行四边形,
∴四边形EDBC是菱形„„„„„„„„10分
17.(9分)如图,四边形ABCD是平行四边形,△AB’C和△ABC关于AC所在的直线对称,AD和B’C相交于点O,连接BB’.
(1)请直接写出图中所有的等腰三角形(不添加字母); (2)求证:△AB’O≌△CDO.
A
17.(9分)如图,在梯形ABCD中,AD∥BC,延长CB到点E,使BE=AD,连接DE交AB于点M.
(1)求证:△AMD≌△BME;
(2)若N是CD的中点,且MN=5,BE=2,求BC的长
.
17.(1)∵AD∥BC,∴∠A=MBE,∠ADM=∠E.„„„„„„„„„„„„„2分 在△AMD和△BME中,
∠A=∠MBE,
∴△AMD≌△BME.„„„„„„„„„„„„„„5分 AD=BE,
∠ADM=E,
(2)∵△AMD≌△BME,∴MD=ME.又ND=NC,∴MN=
EC.„„„„„„„„„„„„„„„„„„„„„„„7分 2
∴EC=2MN=2×5=10.
∴BC=EC-EB=10-2=8.„„„„„„„„„„„„„„„„„„„„„„9分 21.(9分)如图,在梯形ABCD中,AB∥CD,∠BCD=90°,且AB=1,BC=2,tan∠ADC=2.(1)求证:DC=BC;
(2)E是梯形内一点,F是梯形外一点,且∠EDC=∠FBC,DE=BF,试判断△ECF的形状, 并证明你的结论;
(3)在(2)的条件下,当BE:CE=l:2,∠BEC=135°时,求sin∠BFE的值. 2
1、(1)过A作DC的垂线AM交DC于M,则AM=BC=2, 又tan∠ADC=2,∴
AM
22DM1,即DC=BC。 DMDM
(2)等腰直角三角形,证明:∵DE=DF,∠EDC=∠FBC,DC=BC, ∴△DEC≌△BFC ∴CE=CF,∠ECD=∠BCF。 ∴∠ECF=∠BCF+∠BCE=∠ECD+∠BCE=∠BCD=90° 即△ECF是等腰直角三角形
19.(本题满分9分)已知:如图,在△ABC中,D是AC的中点,E是线段BC延长线上一点,过点A作BE的平行线与线段ED的延长线交于点F,连结AE,CF. (1)求证:AFCE;
(2)若ACEF,试判断四边形AFCE是什么样的四边形,并证明你的结论. 19.证明:在△ADF和△CDE中, F AF∥BE,∠FAD∠ECD.
又D是AC的中点,ADCD.D ∠ADF∠CDE,△ADF≌△CDE. AFCE.
C (2)解:若ACEF,则四边形AFCE是矩形.
,四边形AFCE是平行四边形. 由(1),知AF
E
又ACEF,四边形AFCE是矩形.
20.解:⑴交点P所表示的实际意义是:经过2.5小时后,小东与小明在距离B地7.5千米处相遇.⑵设y1kxb,则由题意知
2.5kb7.5m20
,解得
4kb0k
5∴y15x20 ;因为y2是经过原点、P点的直线,所以它的解析式为:y2
x3
(3)当x0时,y120,故AB两地之间的距离为20千米
.