【答案】
分析:(1)四邊形PECF的形狀是正方形,易證四邊形PECF是矩形,由角平分線的性質(zhì)可知:PE=PF,所以四邊形PECF是正方形;
(2)先根據(jù)角平分線及線段垂直平分線的作法作出P點,過點P分別作PE⊥AC、PF⊥CB,垂足為E、F,由全等三角形的判定定理得出Rt△APE≌Rt△BPF,再由全等三角形的性質(zhì)即可判斷出△PAB是等腰直角三角形;
(3)如圖4,在Rt△PAB中,∠APB=90°,PA=PB,PA=m,所以AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/0.png)
PA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/1.png)
,由(2)中的證明過程可知,Rt△AEP≌Rt△BFP,可得AE=BF,CE=CF,所以CA+CB=CE+EA+CB=CE+CF=2CE,又PC=n,所以在正方形PECF中,CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/2.png)
PC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/3.png)
n.所以CA+CB=2CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/4.png)
.進而求出△ABC的周長;
(4)因為∠1=∠2=∠3=∠4=45°,且∠ADC=∠PDB,所以△ADC∽△PDB,故
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/5.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/6.png)
,…①同理可得,△CDB∽△ADP,得到
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/7.png)
,…②又PA=PB,則①+②得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/11.png)
,所以這個值仍不變?yōu)?img src="http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/12.png">.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/images13.png)
解:(1)四邊形PECF的形狀是正方形,理由如下:
過點P分別作PE⊥AC、PF⊥CB,垂足分別為E、F(如圖4)
∵∠ACB=90°,又由作圖可知PE⊥AC、PF⊥CB,
∴四邊形PECF是矩形,
又∵點P在∠ACB的角平分線上,
且PE⊥AC、PF⊥CB,
∴PE=PF,
∴四邊形PECF是正方形;
(2)證明:在Rt△AEP和Rt△BFP中,
∵PE=PF,PA=PB,∠AEP=∠BFP=90°,
∴Rt△AEP≌Rt△BFP,
∴∠APE=∠BPF,
∵∠EPF=90°,從而∠APB=90°.
又因為PA=PB,
∴△PAB是等腰直角三角形;
(3)如圖4,在Rt△PAB中,∠APB=90°,PA=PB,PA=m,
∴AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/13.png)
PA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/14.png)
.
由(2)中的證明過程可知,Rt△AEP≌Rt△BFP,可得AE=BF,CE=CF,
∴CA+CB=CE+EA+CB=CE+CF=2CE,又PC=n,
∴在正方形PECF中,CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/15.png)
PC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/16.png)
n.
∴CA+CB=2CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/17.png)
.
∴△ABC的周長為:AB+BC+CA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/18.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/19.png)
;
(4)當邊AC、BC的長度變化時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/20.png)
的值不變,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/21.png)
.理由如下:
如圖4,∵∠1=∠2=∠3=∠4=45°,且∠ADC=∠PDB,
∴△ADC∽△PDB,故
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/22.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/23.png)
,…①
同理可得,△CDB∽△ADP,得到
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/24.png)
,…②
又PA=PB,則①+②得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/25.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/26.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/28.png)
.
∴這個值仍不變?yōu)?img src="http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193203053107095/SYS201311011932030531070022_DA/29.png">.
點評:本題考查的是相似形綜合題,涉及到角平分線及線段垂直平分線的作法及性質(zhì)、平行線的性質(zhì)、全等三角形的判定與性質(zhì)及三角形的面積公式,涉及面較廣,難度較大.