解:(1)CD=
![](http://thumb.1010pic.com/pic5/latex/53.png)
PB.
理由:过点P分别作PH⊥AC于点H,PF⊥BC于点F,
∴∠PHC=∠PFC=90°,
∵∠BCA=90°,
∴四边形PFCE是矩形,
∴CH=PF,
∵PD=PC,
∴CH=
![](http://thumb.1010pic.com/pic5/latex/13.png)
CD,
在Rt△PBF中,tanB=1,
∴PF=BF,
∴PF=PB•sin45°=
![](http://thumb.1010pic.com/pic5/latex/54.png)
PB,
∴CD=2CH=2PF=2×
![](http://thumb.1010pic.com/pic5/latex/54.png)
PB=
![](http://thumb.1010pic.com/pic5/latex/53.png)
PB;
![](http://thumb.1010pic.com/pic5/upload/201310/528567d2e7124.png)
(2)证明:过点P分别作PH⊥AC于点H,PF⊥BC于点F,
∴∠PHC=∠PFC=90°,
∵∠BCA=90°,
∴四边形PFCE是矩形,
∴CH=PF,
∵PD=PC,
∴CH=
![](http://thumb.1010pic.com/pic5/latex/13.png)
CD,
在Rt△PBF中,tanB=2,
即
![](http://thumb.1010pic.com/pic5/latex/58129.png)
=2,
∴PF=2BF,
由勾股定理得:BP=
![](http://thumb.1010pic.com/pic5/latex/559.png)
BF,PF=
![](http://thumb.1010pic.com/pic5/latex/1285.png)
BP,
∴CH=
![](http://thumb.1010pic.com/pic5/latex/1285.png)
BP,CD=
![](http://thumb.1010pic.com/pic5/latex/33845.png)
BP,
在Rt△ABC中,tanB=2,
同理可得:AC=2BC,
∵AC=AD+CD,
∴2BC=AD+
![](http://thumb.1010pic.com/pic5/latex/33845.png)
BP;
(3)连接BE,
∵点B关于直线CP的对称点为E,
∴CP是线段BE的垂直平分线,
∴CE=CB,PE=PB,
∴∠BCP=∠ECP=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACB=45°,
过点P作PF⊥BC于点F,
设PB=a,
由(2)得:2BC=AD+
![](http://thumb.1010pic.com/pic5/latex/33845.png)
BP,
则BC=1+
![](http://thumb.1010pic.com/pic5/latex/1285.png)
a,
在Rt△CPF中,∠FCP=45°,PF=CF=
![](http://thumb.1010pic.com/pic5/latex/1285.png)
a,
而BF=
![](http://thumb.1010pic.com/pic5/latex/327.png)
BP=
![](http://thumb.1010pic.com/pic5/latex/327.png)
a,
由CF+BF=BC得,
![](http://thumb.1010pic.com/pic5/latex/1285.png)
a+
![](http://thumb.1010pic.com/pic5/latex/327.png)
a=1+
![](http://thumb.1010pic.com/pic5/latex/1285.png)
a,
解得:a=
![](http://thumb.1010pic.com/pic5/latex/559.png)
,
即BP=
![](http://thumb.1010pic.com/pic5/latex/559.png)
,
∴BC=3,AC=2BC=6,AB=3
![](http://thumb.1010pic.com/pic5/latex/559.png)
,AP=2
![](http://thumb.1010pic.com/pic5/latex/559.png)
,CD=4,DE=1,EA=3,
∴BD=
![](http://thumb.1010pic.com/pic5/latex/399067.png)
=5,
过点D作AB的平行线分别交EP于点Q,交CP于点R,
由△EDQ∽△EAP,得ED:EA=DQ:AP=1:3,得DQ=
![](http://thumb.1010pic.com/pic5/latex/10191.png)
,
由△QDM∽△PBM,得DM:BM=QD:PB=2:3,得DM=
![](http://thumb.1010pic.com/pic5/latex/335.png)
BD=2,
由△CDR∽△CAP,得DR:AP=CD:CA=4:6,得DR=
![](http://thumb.1010pic.com/pic5/latex/91044.png)
,
由△NDR∽△NBP,得DN:BN=DR:PB=
![](http://thumb.1010pic.com/pic5/latex/91044.png)
:
![](http://thumb.1010pic.com/pic5/latex/559.png)
=
![](http://thumb.1010pic.com/pic5/latex/304.png)
,得DN=
![](http://thumb.1010pic.com/pic5/latex/4458.png)
BD=
![](http://thumb.1010pic.com/pic5/latex/6482.png)
,
∴NM=DN-DM=
![](http://thumb.1010pic.com/pic5/latex/6482.png)
-2=
![](http://thumb.1010pic.com/pic5/latex/9777.png)
.
分析:(1)首先过点P分别作PH⊥AC于点H,PF⊥BC于点F,又由在△ABC中,∠ACB=90°,易得四边形PFCE是矩形,即可得CH=PF,又由tanB=1,可得∠B=45°,PF=BF,由三角函数可求得PF═
![](http://thumb.1010pic.com/pic5/latex/54.png)
PB,由PC=PD,根据三线合一的性质,可得CD=2CH=2PF,即可求得答案;
(2)证明方法同(1),首先可得四边形PFCE是矩形,CH=PF=
![](http://thumb.1010pic.com/pic5/latex/13.png)
CD,然后由勾股定理得:BP=
![](http://thumb.1010pic.com/pic5/latex/559.png)
BF,PF=
![](http://thumb.1010pic.com/pic5/latex/1285.png)
BP,即可求得答案;
(3)据题意可得CP是线段BE的垂直平分线,即可得CE=CB,PE=PB,则可求得∠BCP=∠ECP=
![](http://thumb.1010pic.com/pic5/latex/13.png)
∠ACB=45°,然后利用勾股定理,借助于方程求解即可BC=3,AC=2BC=6,AB=3
![](http://thumb.1010pic.com/pic5/latex/559.png)
,AP=2
![](http://thumb.1010pic.com/pic5/latex/559.png)
,CD=4,DE=1,EA=3,然后过点D作AB的平行线分别交EP于点Q,交CP于点R,利用相似三角形的对应边成比例,即可求得答案.
点评:此题考查了相似三角形的判定与性质、矩形的判定与性质、勾股定理、等腰三角形的性质、直角三角形的性质以及三角函数等知识.此题综合性很强,难度很大,注意辅助线的作法,注意数形结合思想的应用.