解:(1)∵BC=AD=9,BE=4,
∴CE=9-4=5
∵AF=CE
即:3t=5,
∴t=
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,
∵EH∥DF
∴△DAF∽△EBH,
∴
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=
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即:
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=
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解得:BH=
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;
當(dāng)t=
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時,AF=CE,此時BH=
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;
(2)由EH∥DF得∠AFD=∠BHE,
又∵∠A=∠CBH=90°
∴△EBH∽△DAF,
∴
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即
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=
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∴BH=
當(dāng)點F在點B的左邊時,
即t<4時,BF=12-3t
此時,當(dāng)△BEF∽△BHE時:
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即4
2=(12-3t)×
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解得:t
1=2
此時,當(dāng)△BEF∽△BEH時:有BF=BH,即12-3t=
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解得:t
2=
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當(dāng)點F在點B的右邊時,即t>4時,BF=3t-12
此時,當(dāng)△BEF∽△BHE時:
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即4
2=(3t-12)×
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解得:t
3=2
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+2
(3)①∵EH∥DF
∴△DFE的面積=△DFH的面積=
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FH•AD=
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(12-3t+
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t)×9=54-
②直接寫出C的最小值=13+
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.
分析:(1)在Rt△ABC中,利用勾股定理可求得AB的長,即可得到AD、t的值,從而確定AE的長,由DE=AE-AD即可得解.
(2)若△DEG與△ACB相似,要分兩種情況:①AG:DE=DH:GE,②AH:EG=DH:DE,根據(jù)這些比例線段即可求得t的值.(需注意的是在求DE的表達式時,要分AD>AE和AD<AE兩種情況)
點評:此題考查了勾股定理、軸對稱的性質(zhì)、平行四邊形及梯形的判定和性質(zhì)、解直角三角形、相似三角形等相關(guān)知識,綜合性強,是一道難度較大的壓軸題.