??EDB??AMB.
又
?B??B,
DEBD. ?AMBM115CD=BC=. 24211BC=?10=5, 22?△BDE△BMA.
∴
由AD?AC,AM?BC, 知DM?又
BD??DE5. ?545?28?DE?.
3解析:
18.答案:证明:如图,连接PA
EP是AD的垂直平分线,?PA?PD. ??PDA??PAD,??B??BAD??DAC??CAP
又AD平分?BAC,
??BAD??DAC,??B?CAP
又?APC??BPA,?△PAC△PBA
?PAPC,即PA2?PB?PC ?PBPAPA?PD,?PD2?PB?PC.
解析:
19.答案:1.证明:
AB//FC,
??A??ECF.
又
?AED??CEF,且DE?FE,
?△ADE?△CFE(AAS)2.解:方法一:
AB//FC,
??GBD??GCF,?GDB??F. ?△GBD△GCF
??GBBD. ?GCCF21 ?2?4CF?CF?3.
由(1)得△ADE?△CFE,
?AD?CF?3.
?AB?AD?BD?3?1?4
方法二:如图,取BC的中点H,连接EH
△ADE?△CFE, ?AE?CE
?EH是△ABC的中位线.
?EH//AB,且EH?1AB 2??GBD??GHE,?GDB??GEH. ?△GBD△GHE
?DBGB ?EHGH?12 ?EH2?2?EH?2.
?AB?2EH?4
解析:
20.答案:(1)证明:四边形ABCD为平行四边形,??B??D
AM?BC,AN?CD
??AMB??AND?90?,?△AMB△AND. (2)由△AMB△AND得又
AD?BC,?AMAB,?BAM??DAN ?ANADAMABAMAN ?,??ANBCABBCAM?BC,AD//BC,??MAD??AMB?90? ??B??BAM??MAN??NAD?90?,??B??MAN
?△AMN△BAC,?AMMN. ?ABAC解析:
21.答案:(1)135?;22 (2)△ABC△CED. 证明:
BC?22.EC?2,?AB2BC22ABBC??2,???2,?.又DE2CECEDE2△ABC△CED. ?ABC??CED?135°,? 解析:
22.答案:1. ∵OA=12厘米,OB=6厘米,由题意,得BQ=t,OP=t,∴OQ=6-t. ∴y?1?OP?OQ 21??t??6?t? 21??t2?3t?0?t?6?
2OQOP6?tt,即??,解得t=4.
OBOA612OQOP6?tt②当△POQ∽△BOA时, ,即??,解得t=2.
OAOB1262. ①当△POQ∽△AOB时,
∴当t=4或t=2时,△POQ与△AOB相似.
解析:
23.答案:(1)证明:四边形ABCD是矩形,
?AD?BC,DC?AB,?DAB??B??C??D?90°.由折叠可
得,AP?AB,PO?BO,?PAO??BAO,?APO??B?90°,??APD?90°??CPO??POC,?△OCP△PDA.
(2)解:△OCP与△PDA的面积比为1:4,
?OCOPCP11????.?PD?2OC,PA?2OP,DA?2CP.PDPADA42?CP?4,BC?8.
设AB?x,则AP?BP?x. 在Rt△ADP中,
AD?8,?x2?82?(x?4)2.解x?10.即 AP?x,?D?90°,AD?8,DP?x?4,AB?10.
解析:
24.答案:解:(1)设直线AD的解析式为y?kx?b.将A(,),D(0,1)代入得
453351?4?k?b?,k?,?3?32解得? ?b?1,b?1.????故直线AD的解析式为y?1x?1. 2(2) 直线AD的表达式为y?1x?1. 2令y?0,得x??2.?B(?2,0)
?OB?2.
直线AC的表达式为y??x?3. 令y?0,得x?3.?C(3,0)
?OC?3.
设E(x,1x?1) 2①当E1C?BC时,如图,?BOD??BCE1?90°,?DBO??E1BC.
?△BOD△BCE1
此时点C和点E1的横坐标相同. 将x?3代入y?155x?1,解得y?.?E1(3,). 222②当CE2?AD时,如图,?BOD??BE2C?90°,?DBO??CBE2.
?△BOD△BE2C
乙DBO二乙CBE2, …LBOD-tBE2C.
过点E2作E2F?x轴于点F,则?E2FC??BE2F?90°.
??E2BF??BE2F?90°.
又
?CE2F??BE2F?90°,??E2BF??CE2F.
E2FCF?. BFE2F?△E2BF△CE2F,则
1?E2F2?CF?BF,即(x?1)2?(3?x)(x?2)
2解得x1?2,x2??2 (舍去).
?E2(2,2).
当?EBC?90°时,此情况不存在. 综上所述,点E的坐标为(3,)或(2,2).
52
解析:
25.答案:(1)由题意知AP?2t,DQ?t,QA?6?t, 当QA?AP时,△QAP是等腰直角三角形, 所以6?t?2t,解得t?2. (2)四边形QAPC的面积=SOAC?SAPC?11AQCD?APBC=(36?6t)?6t?36(cm2). 22在P,Q两点移动的过程中,四边形QAPC的面积始终保持不变. (3)分两种情况: ①当AQAP时,△QAP△ABC, ?ABBC则6?t2t?即t?1.2; 126QAAP时,△PAQ△ABC, ?BCAB②当
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