【答案】
分析:(1)過點(diǎn)D作DH⊥BC交BC的延長(zhǎng)線于H,由∠DEB=∠EBH=∠DHB=90°可知四邊形DEBH為矩形,故可得出∠CDH=∠ADE,再由相似三角形的判定定理得出△DCH∽△DAE,由相似三角形的對(duì)應(yīng)邊成比例即可求出CH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/0.png)
AE,故可得出結(jié)論;
(2)過點(diǎn)F作FM⊥DE交DE于M,由題意可得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/3.png)
,故可得出AE及FM的長(zhǎng),由相似三角形的判定定理得出△DCH∽△DAE,由相似三角形的對(duì)應(yīng)邊成比例即可求出CH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/4.png)
AE,根據(jù)四邊形DEBH為矩形得BE=DH;tan∠BDE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/5.png)
,在Rt△DFM′中可得出DM=8,F(xiàn)D=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/6.png)
,設(shè)AG=3a(a>0),AG:FG=3:2,F(xiàn)G=2a,故可得出△DFG∽△AFD,由相似三角形的對(duì)應(yīng)邊成比例可求出FD
2的值,在Rt△AGE中,∠AEG=90°,AG=6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/7.png)
,AE=6,在Rt△FMG中,∠FMG=90°,F(xiàn)G=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/8.png)
,F(xiàn)M=4,GE=6,DE=GE+GM+DM=6+4+8=18,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/9.png)
+BC=DE,BC=DE-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/10.png)
=18-3=15.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/images11.png)
解:(1)如圖1,過點(diǎn)D作DH⊥BC交BC的延長(zhǎng)線于H,
∵∠DEB=∠EBH=∠DHB=90°,
∴四邊形DEBH為矩形,
∴∠ADC=90°,
∴∠CDH+∠EDC=∠ADE+∠EDC=90°,
∴∠CDH=∠ADE,
又∵∠DHC=∠AED=90°,
∴△DCH∽△DAE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/13.png)
,
∴CH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/14.png)
AE,
∵DE=BH而BH=BC+CH=BC+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/15.png)
AE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/16.png)
+BC=DE.
(2)如圖2,過點(diǎn)F作FM⊥DE交DE于M,
∴∠FMG=90°,
又∵∠AED=90°,
∴∠FMG=∠AED,而∠FGM=∠AGE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/19.png)
,
∵AE=6,
∴FM=4,
由(1)知,△DCH∽△DAE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/22.png)
,而由四邊形DEBH為矩形得BE=DH,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/23.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/24.png)
,
∴tan∠BDE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/25.png)
,
在Rt△DFM′中,∠FMD=90°,tan∠FMD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/26.png)
,F(xiàn)M=4,
∴DM=8,F(xiàn)D=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/27.png)
,
設(shè)AG=3a(a>0),
∵AG:FG=3:2,
∴FG=2a,
∵∠DFG=∠AFD,∠BDE=∠DAC,
∴△DFG∽△AFD,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/28.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/29.png)
,
∴FD
2=FA•FG,
∴(4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/30.png)
)
2=(3a+2a)•2a,
∴a=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/31.png)
,
∴FG=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/32.png)
,AG=6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/33.png)
,
在Rt△AGE中,∠AEG=90°,AG=6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/34.png)
,AE=6,
∴GM=4,
在Rt△FMG中,∠FMG=90°,F(xiàn)G=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/35.png)
,F(xiàn)M=4,
∴GE=6,
∴DE=GE+GM+DM=6+4+8=18,
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/36.png)
+BC=DE,
∴BC=DE-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190927753991119/SYS201311011909277539911027_DA/37.png)
=18-3=15.
點(diǎn)評(píng):本題考查了相似形綜合題,涉及到相似三角形的判定與性質(zhì)、解直角三角形的知識(shí),難度較大.