【答案】
分析:(1)過點(diǎn)D作DH⊥BC交BC的延長線于H,由∠DEB=∠EBH=∠DHB=90°可知四邊形DEBH為矩形,故可得出∠CDH=∠ADE,再由相似三角形的判定定理得出△DCH∽△DAE,由相似三角形的對應(yīng)邊成比例即可求出CH=
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AE,故可得出結(jié)論;
(2)過點(diǎn)F作FM⊥DE交DE于M,由題意可得
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=
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=
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,故可得出AE及FM的長,由相似三角形的判定定理得出△DCH∽△DAE,由相似三角形的對應(yīng)邊成比例即可求出CH=
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AE,根據(jù)四邊形DEBH為矩形得BE=DH;tan∠BDE=
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,在Rt△DFM′中可得出DM=8,F(xiàn)D=4
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,設(shè)AG=3a(a>0),AG:FG=3:2,F(xiàn)G=2a,故可得出△DFG∽△AFD,由相似三角形的對應(yīng)邊成比例可求出FD
2的值,在Rt△AGE中,∠AEG=90°,AG=6
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,AE=6,在Rt△FMG中,∠FMG=90°,F(xiàn)G=4
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,F(xiàn)M=4,GE=6,DE=GE+GM+DM=6+4+8=18,
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+BC=DE,BC=DE-
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=18-3=15.
解答:
解:(1)如圖1,過點(diǎn)D作DH⊥BC交BC的延長線于H,
∵∠DEB=∠EBH=∠DHB=90°,
∴四邊形DEBH為矩形,
∴∠ADC=90°,
∴∠CDH+∠EDC=∠ADE+∠EDC=90°,
∴∠CDH=∠ADE,
又∵∠DHC=∠AED=90°,
∴△DCH∽△DAE,
∴
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=
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=
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,
∴CH=
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AE,
∵DE=BH而BH=BC+CH=BC+
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AE,
∴
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+BC=DE.
(2)如圖2,過點(diǎn)F作FM⊥DE交DE于M,
∴∠FMG=90°,
又∵∠AED=90°,
∴∠FMG=∠AED,而∠FGM=∠AGE,
∴
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=
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=
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,
∵AE=6,
∴FM=4,
由(1)知,△DCH∽△DAE,
∴
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=
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=
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,而由四邊形DEBH為矩形得BE=DH,
∴
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=
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,
∴tan∠BDE=
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,
在Rt△DFM′中,∠FMD=90°,tan∠FMD=
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,F(xiàn)M=4,
∴DM=8,F(xiàn)D=4
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,
設(shè)AG=3a(a>0),
∵AG:FG=3:2,
∴FG=2a,
∵∠DFG=∠AFD,∠BDE=∠DAC,
∴△DFG∽△AFD,
∴
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=
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,
∴FD
2=FA•FG,
∴(4
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)
2=(3a+2a)•2a,
∴a=2
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,
∴FG=4
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,AG=6
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,
在Rt△AGE中,∠AEG=90°,AG=6
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,AE=6,
∴GM=4,
在Rt△FMG中,∠FMG=90°,F(xiàn)G=4
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,F(xiàn)M=4,
∴GE=6,
∴DE=GE+GM+DM=6+4+8=18,
∵
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+BC=DE,
∴BC=DE-
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=18-3=15.
點(diǎn)評:本題考查了相似形綜合題,涉及到相似三角形的判定與性質(zhì)、解直角三角形的知識,難度較大.