【答案】
分析:(1)由已知條件可以得出AD=AP,∠DAP=∠BAC=60°,∠ADM=∠APN=60°,從而得出∠DAM=∠PAN,可以得出△ADM≌△APN,就可以得出結(jié)論.
(2)①由已知條件可以得出△BPM∽△CAP,可以得出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/0.png)
,由已知條件可以建立方程求出BP的值.
②四邊形AMPN的面積就是四邊形ADPE與△ABC重疊部分的面積,由△ADM≌△APN,S
△ADM=S
△APN,可以得出重合部分的面積就是△ADP的面積.
③連接PG,若∠DAB=15°,由∠DAP=60°可以得出∠PAG=45°.由已知條件可以得出四邊形ADPE是菱形,就有DO垂直平分AP,得到GP=AG,就有∠PAG=∠APG=45°,得出∠PGA=90°,設(shè)BG=t,在Rt△BPG中∠APG=60°,就可以求出BP=2t,PG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/1.png)
t,從而求得t的值,即可以求出結(jié)論.以DG、GH、HE這三條線段為邊構(gòu)成的三角形為直角三角形,由已知條件可知四邊形ADPE為菱形,可以得到∠ADO=∠AEH=30°,根據(jù)∠DAB=15°,可以求出∠AGO=45°,∠HAO=15°,∠EAH=45°.設(shè)AO=a,則AD=AE=2a,OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/2.png)
a,得到DG=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/3.png)
-1)a,由∠DAB=15°,可以求出∠DHA=∠DAH=75°,求得GH=(3-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/4.png)
)a,HE=2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/5.png)
-1)a,最后由勾股定理的逆定理就可以得出結(jié)論.
解答:解:(1)證明:∵△ABC、△APD和△APE是等邊三角形,
∴AD=AP,∠DAP=∠BAC=60°,∠ADM=∠APN=60°,
∴∠DAM=∠PAN.
在△ADM和△APN中,
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/6.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/images7.png)
∴△ADM≌△APN,
∴AM=AN.
(2)①∵△ABC、△ADP是等邊三角形,
∴∠B=∠C=∠DAP=∠BAC=60°,
∴∠DAM=∠PAC,
∵∠ADM=∠B,∠DMA=∠BMP,
∴180-∠ADM-∠DMA=180-∠B-∠BMP,
∴∠DAM=∠BPM,
∴∠BPM=∠NAP,
∴△BPM∽△CAP,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/7.png)
,
∵BM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/8.png)
,AC=2,CP=2-x,
∴4x
2-8x+3=0,
解得x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/9.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/10.png)
.
②∵四邊形AMPN的面積即為四邊形ADPE與△ABC重疊部分的面積.
∵△ADM≌△APN,
∴S
△ADM=S
△APN,
∴S
四邊形AMPN=S
△APM+S
△APN=S
△AMP+S
△ADM=S
△ADP.
過點(diǎn)P作PS⊥AB,垂足為S,
在Rt△BPS中,∵∠B=60°,BP=x,
PS=BPsin60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/11.png)
x,BS=BPcos60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/12.png)
x,
∵AB=2,
∴AS=AB-BS=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/13.png)
x,
∴AP
2=AS
2+PS
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/14.png)
=x
2-2x+4.
取AP的中點(diǎn)T,連接DT,在等邊三角形ADP中,DT⊥AP,
∴S△ADP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/15.png)
AP.DT=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/16.png)
AP×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/18.png)
,
∴S=S四邊形AMPN=S△ADP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/20.png)
(0<x<2),
∴當(dāng)x=1時(shí),S的最小值是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/21.png)
.
③連接PG,若∠DAB=15°,
∵∠DAP=60°,
∴∠PAG=45°.
∵△APD和△APE是等邊三角形,
∴四邊形ADPE是菱形,
∴DO垂直平分AP,
∴GP=AG,
∴∠PAG=∠APG=45°,
∴∠PGA=90°.
設(shè)BG=t,在Rt△BPG中,∠ABP=60°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/images23.png)
∴BP=2t,PG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/22.png)
t,
∴AG=PG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/23.png)
t,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/24.png)
t+t=2,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/25.png)
-1,
∴BP=2t=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/26.png)
-2.
∴當(dāng)BP=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/27.png)
-2時(shí),∠BAD=15°.
猜想:以DG、GH、HE這三條線段為邊構(gòu)成的三角形為直角三角形.
設(shè)DE交AP于點(diǎn)O,
∵△APD和△APE是等邊三角形,
∴AD=DP=AP=PE=EA,
∴四邊形ADPE為菱形,
∴AO⊥DE,∠ADO=∠AEH=30°.
∵∠DAB=15°,
∴∠GAO=45°,
∴∠AGO=45°,∠HAO=15°,
∴∠EAH=45°.
設(shè)AO=a,則AD=AE=2a,GO=AO=a,OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/28.png)
a.
∴DG=DO-GO=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/29.png)
-1)a.
∵∠DAB=15°,∠BAC=60°,∠ADO=30°,
∴∠DHA=∠DAH=75°.
∴DH=AD=2a,
∴GH=DH-DG=2a-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/30.png)
-1)a=(3-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/31.png)
)a.
HE=DE-DH=2DO-DH=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/32.png)
a-2a.
∵DG
2+GH
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/33.png)
,
HE
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/34.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193327722687392/SYS201311011933277226873023_DA/35.png)
.
∴DG
2+GH
2=HE
2,
∴以DG、GH、HE這三條線段為邊構(gòu)成的三角形為直角三角形.
點(diǎn)評(píng):本題考查了等邊三角形的性質(zhì),全等三角形的判定與性質(zhì)的運(yùn)用,相似三角形的判定與性質(zhì)以及勾股定理的運(yùn)用.本題的綜合性較強(qiáng)在解答時(shí)要注意解答問題的突破口,這也是解答問題的關(guān)鍵.