(2010•高淳县一模)如图1,在正方形ABCD中,对角线AC与BD相交于点O,AF平分∠BAC,交BD于点F.
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(2010•高淳县一模)如图1,在正方形ABCD中,对角线AC与BD相交于点O,AF平分∠BAC,交BD于点F.
(1)求证:DF=AD;
(2)过点F作FH⊥AB,垂足为点H,求证:FH+
(1)求证:DF=AD;
(2)过点F作FH⊥AB,垂足为点H,求证:FH+
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![(2010•高淳县一模)如图1,在正方形ABCD中,对角线AC与BD相交于点O,AF平分∠BAC,交BD于点F.](/uploads/image/z/434780-44-0.jpg?t=%EF%BC%882010%E2%80%A2%E9%AB%98%E6%B7%B3%E5%8E%BF%E4%B8%80%E6%A8%A1%EF%BC%89%E5%A6%82%E5%9B%BE1%EF%BC%8C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%EF%BC%8C%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E4%B8%8EBD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9O%EF%BC%8CAF%E5%B9%B3%E5%88%86%E2%88%A0BAC%EF%BC%8C%E4%BA%A4BD%E4%BA%8E%E7%82%B9F%EF%BC%8E)
(1)证明:∵正方形ABCD,
∴∠DAC=∠ABD=45°,![](http://img.wesiedu.com/upload/6/28/6282ca1b3df0ce933636c95b989da0e8.jpg)
∵AF平分∠BAC,
∴∠CAF=∠BAF,
而∠DAF=∠DAC+∠FAC,∠DFA=∠ABD+∠BAF,
∴∠DAF=∠DFA,
∴DF=AD;
(2)证明:∵正方形ABCD,
∴FO⊥AC,
1
2AC=OD,
∵AF平分∠BAC,FH⊥AB,![](http://img.wesiedu.com/upload/0/4c/04ca8bd762421afd2f0b760031f06826.jpg)
∴FH=FO,
∴FH+
1
2AC=FO+OD=DF=AD,
即FH+
1
2AC=AD.
(3)猜想:F1H1+
1
2A1C1=AD.
理由:∵AD=CD,∠ADC=∠A1DC1,
∴∠A1DA=∠C1DC,
∴△A1AD≌△C1CD,
∴△A1C1D是等腰直角三角形,
∵A1F1平分∠BA1C1,
∴∠BA1F1=∠F1A1C1
而∠DA1F1=45°+∠F1A1C1,∠DF1A1=45°+∠BA1F1,
∴∠DA1F1=∠DF1A1,
∴A1D=DF1,
∴
1
2A1C1=
2
2A1D=
2
2DF1,
又∵在等腰直角三角形F1H1B中,F1H1=
2
2F1B,
∴F1H1+
1
2A1C1=
2
2F1B+
∴∠DAC=∠ABD=45°,
![](http://img.wesiedu.com/upload/6/28/6282ca1b3df0ce933636c95b989da0e8.jpg)
∵AF平分∠BAC,
∴∠CAF=∠BAF,
而∠DAF=∠DAC+∠FAC,∠DFA=∠ABD+∠BAF,
∴∠DAF=∠DFA,
∴DF=AD;
(2)证明:∵正方形ABCD,
∴FO⊥AC,
1
2AC=OD,
∵AF平分∠BAC,FH⊥AB,
![](http://img.wesiedu.com/upload/0/4c/04ca8bd762421afd2f0b760031f06826.jpg)
∴FH=FO,
∴FH+
1
2AC=FO+OD=DF=AD,
即FH+
1
2AC=AD.
(3)猜想:F1H1+
1
2A1C1=AD.
理由:∵AD=CD,∠ADC=∠A1DC1,
∴∠A1DA=∠C1DC,
∴△A1AD≌△C1CD,
∴△A1C1D是等腰直角三角形,
∵A1F1平分∠BA1C1,
∴∠BA1F1=∠F1A1C1
而∠DA1F1=45°+∠F1A1C1,∠DF1A1=45°+∠BA1F1,
∴∠DA1F1=∠DF1A1,
∴A1D=DF1,
∴
1
2A1C1=
2
2A1D=
2
2DF1,
又∵在等腰直角三角形F1H1B中,F1H1=
2
2F1B,
∴F1H1+
1
2A1C1=
2
2F1B+
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