解:(1)①∵∠APQ+∠CPQ=∠B+∠BAP,∠APQ=∠ABC,
∴∠BAP=∠CQP.
又∵AB=AC,∴∠B=∠C.
∴△CPQ∽△BAP.
∴
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.
∵AB=AC=5,BC=8,BP=6,CP=8-6=2,
∴
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,
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.
②若點P在線段CB上,由(1)知
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,
∵BP=x,BC=8,∴CP=BC-BP=8-x,
又∵CQ=y,AB=5,∴
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,即
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.
故所求的函數(shù)關(guān)系式為
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,(0<x<8).
若點P在線段CB的延長線上,如圖.
∵∠APQ=∠APB+∠CPQ,
∠ABC=∠APB+∠PAB,∠APQ=∠ABC,
∴∠CPQ=∠PAB.
又∵∠ABP=180°-∠ABC,∠PCQ=180°-∠ACB,∠ABC=∠ACB,
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∴∠ABP=∠PCQ.∴△QCP∽△PBA.∴
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.
∵BP=x,CP=BC+BP=8+x,AB=5,CQ=y,
∴
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,即
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(x≥8).
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(2)①當點P在線段BC上,
∵∠APQ=90°,
∴∠APB+∠QPC=90°,
∵∠PAB+∠APB=90°,
∴∠PAB=∠QPC,
∵∠B=∠C=90°,
∴△ABP∽△PCQ,
∴AB:PC=BP:CQ,
即5:(5-BP)=BP:1,
解得:
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,或
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,
②當點P在線段BC的延長線上,則點Q在線段DC的延長線上,
同理可得:△ABP∽△PCQ,
∴AB:PC=BP:CQ,
∴5:(BP-5)=BP:1,
解得:
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,
③當點P在線段CB的延長線上,則點Q在線段DC的延長線上,
同理可得:△ABP∽△PCQ,
∴AB:PC=BP:CQ,
∴5:(BP+5)=BP:1,
解得:
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.
分析:(1)①求線段CQ的長,根據(jù)已知條件AB=AC,∠APQ=∠ABC知道,可以先證明△QCP∽△PBA,由比例關(guān)系式得出;
②要求y與x之間的函數(shù)關(guān)系式,函數(shù)的定義域,因為BP在線段CB上,或在CB的延長線上,根據(jù)實際情況證明△QCP∽△ABP,求出比例關(guān)系式得出
(2)要求線段BP的長,先證明△BAP∽△CPQ得出比例式,再利用圖形間的“和差“關(guān)系求解.
點評:本題結(jié)合三角形,正方形的性質(zhì)考查二次函數(shù)的綜合應(yīng)用,根據(jù)相似三角形的性質(zhì),利用圖形間的“和差“關(guān)系求解.