解:(1)∵直線y=kx+b與x軸交于點(diǎn)A(8,0),與y軸交于點(diǎn)B(0,16),
∴
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,
解得
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,
所以,直線AB的解析式為y=-2x+16,
聯(lián)立
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,
解得
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,
所以,C點(diǎn)坐標(biāo)為(
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,
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);
(2)根據(jù)題意,點(diǎn)D、E的縱坐標(biāo)都是t,
所以,-2x+16=t,
解得x=
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,
所以,點(diǎn)D(t,t),E(
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,t),
DE=|
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-t|,
∵點(diǎn)F在x軸上,
∴|
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-t|=t,
即
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-t=t或
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-t=-t,
解得t=
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或t=16,
所以,t的值為
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,16;
(3)①PE∥AF時(shí),點(diǎn)F在x軸上,根據(jù)(2)的結(jié)論,
t=
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或16,
當(dāng)t=16時(shí),P、B、E三點(diǎn)重合,以A,E,P,F(xiàn)為頂點(diǎn)的是三角形,不符合題意舍去,
所以,t=
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;
②PF∥AE時(shí),點(diǎn)D在點(diǎn)E的左邊,
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∵D(t,t),E(
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,t),
∴DE=
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-t=
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,
點(diǎn)F的縱坐標(biāo)為:t-
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=
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,
∴點(diǎn)F(t,
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),
設(shè)直線PF的解析式為y=ex+f,
則
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,
解得
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,
所以,直線PF的解析式為y=
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x+t,
∵PF∥AE,
∴
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=-2,
解得t=
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;
③AP∥EF時(shí),(i)若點(diǎn)P在y軸正半軸,則DE=t-

=
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,
點(diǎn)F的縱坐標(biāo)為t-
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=
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,
∴點(diǎn)F的坐標(biāo)為(t,
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),
設(shè)直線EF的解析式為y=cx+d,則
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,
解得
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,
∴直線EF的解析式為y=-x+
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,
又∵A(8,0),P(0,t),
∴直線AP的解析式為y=-
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+t,
∵AP∥EF,
∴-
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=-1,
解得t=8,
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(ii)若點(diǎn)P在y軸負(fù)半軸,則DE=
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-t=
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,
點(diǎn)F的縱坐標(biāo)為t-
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=
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,
∴點(diǎn)F的坐標(biāo)為(t,

),
設(shè)直線EF的解析式為y=mx+n,則
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,
解得
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,
∴直線EF的解析式為y=x+
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,
又∵A(8,0),P(0,t),
∴直線AP的解析式為y=-
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+t,
∵AP∥EF,
∴-
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=1,
解得t=-8,
綜上所述,t的值為
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,
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,8,-8.
分析:(1)把點(diǎn)A、B的坐標(biāo)代入直線y=kx+b得到關(guān)于k、b二元一次方程組,求解得到k、b的值,即可得解,聯(lián)立兩直線解析式,求解即可得到點(diǎn)C的坐標(biāo);
(2)利用直線解析式表示出點(diǎn)D、E的坐標(biāo),然后求出DE的長度,再根據(jù)點(diǎn)F在x軸上,DE=DF列式計(jì)算即可得解;
(3)根據(jù)梯形的底邊平行,分①PE∥AF時(shí),點(diǎn)F在x軸上,根據(jù)(2)的結(jié)論解答,②PF∥AE時(shí),先根據(jù)點(diǎn)D、E的坐標(biāo)求出DE的長度,然后表示出點(diǎn)F的坐標(biāo),再利用待定系數(shù)法求出直線PF的解析式,然后根據(jù)平行直線的解析式的k值相等列式求解即可得到t的值;③AP∥EF時(shí),分點(diǎn)P在y軸正半軸與負(fù)半軸兩種情況求出DE的長度,然后表示出點(diǎn)F的坐標(biāo),再利用待定系數(shù)法求出直線AP、EF的解析式,然后根據(jù)平行直線的解析式的k值相等列式求解即可得到t的值.
點(diǎn)評(píng):本題是一次函數(shù)的綜合題型,主要涉及到待定系數(shù)法求一次函數(shù)解析式,聯(lián)立兩直線解析式求交點(diǎn)坐標(biāo),等腰直角三角形的性質(zhì),以及梯形的兩底邊互相平行,(3)求解思路比較復(fù)雜,且運(yùn)算量較大,要分情況討論求解.