(1)證明:證法一:
如圖①,∵BA⊥AM,MN⊥AC,
∴∠BAM=ANM=90°,
∴∠PAQ+∠MAN=∠MAN+∠AMN=90°,
∴∠PAQ=∠AMN,
∵PQ⊥AB MN⊥AC,
∴∠PQA=∠ANM=90°,
∴AQ=MN,
∴△AQP≌△MNA(ASA)
∵AN=PQ AM=AP,
∴∠AMB=∠APM
∵∠APM=∠BPC,∠BPC+∠PBC=90°,∠AMB+∠ABM=90°
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∴∠ABM=∠PBC
∵PQ⊥AB,PC⊥BC
∴PQ=PC(角平分線的性質(zhì)),
∴PC=AN;
證法二:
如圖①,∵BA⊥AM,MN⊥AC,
∴∠BAM=ANM=90°
∴∠PAQ+∠MAN=∠MAN+∠AMN=90°
∴∠PAQ=∠AMN
∵PQ⊥AB,
∴∠AQP=90°=∠ANM
∵AQ=MN,
∴△PQA≌△ANM(ASA)
∴AP=AM,PQ=AN,
∴∠APM=∠AMP
∵∠AQP+∠BAM=180°,
∴PQ∥MA
∴∠QPB=∠AMP
∵∠APM=∠BPC,
∴∠QPB=∠BPC
∵∠BQP=∠BCP=90°,BP=BP
∴△BPQ≌△BPC(AAS)
∴PQ=PC,
∴PC=AN.
(2)解:解法一:
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如圖②,∵NP=2 PC=3,
∴由(1)知PC=AN=3
∴AP=NC=5 AC=8,
∴AM=AP=5
∴AQ=MN=
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=4
∵∠PAQ=∠AMN,∠ACB=∠ANM=90°
∴∠ABC=∠MAN
∴tan∠ABC=tan∠MAN=
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=
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∵tan∠ABC=
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,
∴BC=6
∵NE∥KC,
∴∠PEN=∠PKC,
又∵∠ENP=∠KCP
∴△PNE∽△PCK,
∴
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=
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,
∵CK:CF=2:3,
設(shè)CK=2k,則CF=3k
∴
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=
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,NE=
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k.
過N作NT∥EF交CF于T,則四邊形NTFE是平行四邊形
∴NE=TF=
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k,
∴CT=CF-TF=3k-
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k=
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k
∵EF⊥PM,
∴∠BFH+∠HBF=90°=∠BPC+∠HBF,
∴∠BPC=∠BFH
∵EF∥NT,
∴∠NTC=∠BFH=∠BPC
tan∠NTC=tan∠BPC=
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=2,
∴tan∠NTC=
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=2,
∴CT=
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k=
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,
∴k=
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,
∴CK=2×
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=3,BK=BC-CK=3
∵∠PKC+∠DKB=∠ABC+∠BDK,∠DKE=∠ABC,
∴∠BDK=∠PKC,
tan∠PKC=
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=1,
∴tan∠BDK=1.
過K作KG⊥BD于G
∵tan∠BDK=1,tan∠ABC=
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,
∴設(shè)GK=4n,則BG=3n,GD=4n
∴BK=5n=3,
∴n=
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,
∴BD=4n+3n=7n=
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∵AB=
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=10,AQ=4,
∴BQ=AB-AQ=6
∴DQ=BQ-BD=6-
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=
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.
解法二:
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如圖③,∵NP=2,PC=3,
∴由(1)知PC=AN=3
∴AP=NC=5,AC=8,
∴AM=AP=5
∴AQ=MN=
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=4
∵NM∥BC,
∴∠NMP=∠PBC
又∵∠MNP=∠BCP,
∴△MNP∽△BCP
∴
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=
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,
∴
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=
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BC=6
作ER⊥CF于R,則四邊形NERC是矩形
∴ER=NC=5,NE=CR
∵∠BHE=∠BCR=90°
∴∠EFR=90°-∠HBF∠BPC=90°-∠HBF
∴∠EFR=∠BPC,
∴tan∠EFR=tan∠BPC,
∴
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=
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,即
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=
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∴RF=
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,
∵NE∥KC,
∴∠NEP=∠PKC
又∵∠ENP=∠KCP,
∴△NEP∽△CKP,
∴
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=
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=
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∵CK:CF=2:3,設(shè)CK=2k,CF=3k
∴NE=CR=
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k,CR=CF-RF=3k-
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,
∴3k-
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=
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k
∴k=
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,
∴CK=3 CR=2
∴BK=3
在CF的延長(zhǎng)線上取點(diǎn)G,使∠EGR=∠ABC,
∴tan∠EGR=tan∠ABC
∴
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=
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=
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,
∴RG=
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ER=
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,EG=
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=
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,KG=KC+CR+RG=
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,
∵∠DKE+∠EKC=∠ABC+∠BDK,∠ABC=∠DKE,
∴∠BDK=∠EKC,
∴△BDK∽△GKE,
∴
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=
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∴BD•EG=BK•KG,
∴∠BDK=∠EKC,
∴△BDK∽△GKE,
∴BD=
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∵AB=
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=10,AQ=4,
∴BQ=AB-AQ=6
∴DQ=BQ-BD=6-
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=
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解法三:
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如圖④,∵NP=2,PC=3,
∴由(1)知PC=AN=3
∴AP=NC=5,AC=8,
∴AM=AP=5
∴AQ=MN=
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=4
∵NM∥BC,
∴∠EMH=∠PBC∠PEN=∠PKC
又∵∠PNE=∠PCK,
∴△PNE∽△PCK,△PNM∽△PCB
∴
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=
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,
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=
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,
∵CK:CF=2:3,
設(shè)CK=2k,CF=3k
∴
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=
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,
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=
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,
∴NE=
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k,BC=6
∴BF=6+3k,ME=MN-NE=4-
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k
tan∠ABC=
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=
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,BP=
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=3
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∴sin∠EMH=sin∠PBC=
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=
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∵EF⊥PM,
∴FH=BFsin∠PBC=
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(6+3k)
EH=EMsin∠EMH=
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(4-
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k)
∴tan∠REF=tan∠PBC=
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,
∵tan∠REF=
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∴RF=
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∴EF=
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=
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,
∵EH+FH=EF
∴
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(4-
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k)+
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(6+3k)=
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,
∴k=
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∴CK=2×
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=3,BK=BC-CK=3
∵∠PKC+∠DKE=∠ABC+∠BDK∠DKE=∠ABC,
∴∠BDK=∠PKC
∵tan∠PKC=1,
∴tan∠BDK=1,
過K作KG⊥BD于G
∵tan∠BDK=1,tan∠ABC=
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∴設(shè)GK=4n,則BG=3n,GD=4n
∴BK=5n=3,
∴n=
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,
∴BD=4n+3n=7n=
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∵AB=
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=10,AQ=4,
∴BQ=AB-AQ=6,
∴DQ=BQ-BD=6-
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=
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.
分析:(1)要點(diǎn)是確定一對(duì)全等三角形△AQP≌△MNA,得到AN=PQ;然后推出BP為角平分線,利用角平分線的性質(zhì)得到PC=PQ;從而得到PC=AN;
(2)要點(diǎn)是按照已知條件,求出線段KC的長(zhǎng)度,從而確定△PKC是等腰直角三角形;然后在△BDK中,解直角三角形即可求得BD、DQ的長(zhǎng)度.
點(diǎn)評(píng):本題是幾何綜合題,綜合考查了相似三角形、全等三角形、勾股定理、解直角三角形、角平分線性質(zhì)、平行四邊形、矩形等重要知識(shí)點(diǎn).題干中給出的條件較多,圖形復(fù)雜,難度較大,對(duì)考生能力要求較高;解題時(shí),需要認(rèn)真分析題意,以圖形的相似、圖形的全等為主線尋找解題思路.解答中提供了多種解題方法,可以開拓思路,希望同學(xué)們認(rèn)真研究學(xué)習(xí).