如圖,四邊形ABCD為正方形,△BEF為等腰直角三角形(∠BFE=90°,點(diǎn)B、E、F按逆時(shí)針排列),點(diǎn)P為DE的中點(diǎn),連PC,PF
(1)如圖①,點(diǎn)E在BC上,則線段PC、PF的數(shù)量關(guān)系為______,位置關(guān)系為______(不證明).
(2)如圖②,將△BEF繞點(diǎn)B順時(shí)針旋轉(zhuǎn)a(O<a<45°),則線段PC,PF有何數(shù)量關(guān)系和位置關(guān)系?請(qǐng)寫出你的結(jié)論,并證明.
(3)如圖③,△AEF為等腰直角三角形,且∠AEF=90°,△AEF繞點(diǎn)A逆時(shí)針旋轉(zhuǎn)過程中,能使點(diǎn)F落在BC上,且AB平分EF,直接寫出AE的值是______
【答案】
分析:(1)由∠BFE=90°,點(diǎn)P為DE的中點(diǎn),根據(jù)直角三角形斜邊上的中線等于斜邊的一半得到PF=PD=PE,PC=PD=PE,則PC=PF,又∠FPE=2∠FDP,∠CPE=2∠PDC,得到∠FPC=2∠FDC=90°,所以PC=PF,PC⊥PF.
(2)延長(zhǎng)FP至G使PG=PF,連DG,GC,F(xiàn)C,延長(zhǎng)EF交BD于N,易得△PDG≌△PEF,得DG=EF=BF,得∠PEF=∠PDG,EN∥DG,可得
∠FBC=∠GDC,證得△BFC≌△DGC,則FC=CG,∠BCF=∠DCG.得∠FCG=∠BCD=90°.即有PC⊥PF,PF=PC.
(3)設(shè)AE=2x,則PE=PF=x,AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/0.png)
x,PB=AB-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/1.png)
x,由Rt△AEP∽R(shí)t△FBP,得到
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/3.png)
,解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/4.png)
.得到AE=2x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/5.png)
AB.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/images6.png)
解:(1)∵∠BFE=90°,點(diǎn)P為DE的中點(diǎn)
∴PF=PD=PE,
同理可得PC=PD=PE,
∴PC=PF,
又∵∠FPE=2∠FDP,∠CPE=2∠PDC,
∴∠FPC=2∠FDC=90°,
所以PC=PF,PC⊥PF.
(2)PC⊥PF,PF=PC.理由如下:
延長(zhǎng)FP至G使PG=PF,連DG,GC,F(xiàn)C,延長(zhǎng)EF交BD于N,如圖,
∵點(diǎn)P為DE的中點(diǎn),
∴△PDG≌△PEF,
∴DG=EF=BF.
∴∠PEF=∠PDG,
∴EN∥DG,
∴∠BNE=∠BDG=45°+∠CDG=90°-∠NBF=90°-(45°-∠FBC)
∴∠FBC=∠GDC,
∴△BFC≌△DGC,
∴FC=CG,∠BCF=∠DCG.
∴∠FCG=∠BCD=90°.
∴△FCG為等腰直角三角形,
∵PF=PG,
∴PC⊥PF,PF=PC.
(3)設(shè)AE=2x,則PE=PF=x,AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/6.png)
x,PB=AB-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/7.png)
x,
∵Rt△AEP∽R(shí)t△FBP,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/9.png)
,
∴x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/10.png)
AB.
∴AE=2x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/11.png)
AB.
故答案為PC=PF,PC⊥PF;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232918660757896/SYS201310212329186607578023_DA/12.png)
AB.
點(diǎn)評(píng):本題考查了旋轉(zhuǎn)的性質(zhì):旋轉(zhuǎn)前后的兩個(gè)圖形全等,對(duì)應(yīng)點(diǎn)與旋轉(zhuǎn)中心的連線段的夾角等于旋轉(zhuǎn)角,對(duì)應(yīng)點(diǎn)到旋轉(zhuǎn)中心的距離相等.也考查了正方形的性質(zhì)和三角形全等的判定與性質(zhì)以及三角形相似的性質(zhì).