解:(1)在Rt△ABC中,
∵∠A=90°,AB=6,AC=8,
∴BC=
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=10.
∵∠DHB=∠A=90°,∠B=∠B.
∴△BHD∽△BAC,
∴
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=
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,
∴DH=
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•AC=
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×8=
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(2)∵QR∥AB,
∴∠QRC=∠A=90°.
∵∠C=∠C,
∴△RQC∽△ABC,
∴
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=
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,∴
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=
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,
即y關(guān)于x的函數(shù)關(guān)系式為:y=
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x+6.
(3)存在,分三種情況:
①當(dāng)PQ=PR時,過點P作PM⊥QR于M,則QM=RM.
∵∠1+∠2=90°,∠C+∠2=90°,
∴∠1=∠C.
∴cos∠1=cosC=
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=
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,
∴
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=
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,
∴
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=
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,
∴x=
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.
②當(dāng)PQ=RQ時,-
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x+6=
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,
∴x=6.
③作EM⊥BC,RN⊥EM,
∴EM∥PQ,
當(dāng)PR=QR時,則R為PQ中垂線上的點,
∴EN=MN,
∴ER=RC,
∴點R為EC的中點,
∴CR=
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CE=
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AC=2.
∵tanC=
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=
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,
∴
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=
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,
∴x=
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.
綜上所述,當(dāng)x為
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或6或
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時,△PQR為等腰三角形.
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分析:(1)根據(jù)三角形相似的判定定理求出△BHD∽△BAC,根據(jù)相似三角形的性質(zhì)求出DH的長;
(2)根據(jù)△RQC∽△ABC,根據(jù)三角形的相似比求出y關(guān)于x的函數(shù)關(guān)系式;
(3)畫出圖形,根據(jù)圖形進行討論:
①當(dāng)PQ=PR時,過點P作PM⊥QR于M,則QM=RM.由于∠1+∠2=90°,∠C+∠2=90°,∴∠1=∠C.
∴cos∠1=cosC=
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=
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,∴
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=
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,即可求出x的值;
②當(dāng)PQ=RQ時,-
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x+6=
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,x=6;
③當(dāng)PR=QR時,則R為PQ中垂線上的點,于是點R為EC的中點,故CR=
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CE=
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AC=2.由于tanC=
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=
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,x=
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.
點評:本題很復(fù)雜,把一次函數(shù)與三角形的知識相結(jié)合,使題目的綜合性加強,提高了難度,解答此題的關(guān)鍵是根據(jù)題意畫出圖形,用數(shù)形結(jié)合的方法解答.