【答案】
分析:(1)由F為BD中點(diǎn),DE⊥AB,根據(jù)直角三角形斜邊上的中線等于斜邊的一半,即可得到CF=EF;
(2)過點(diǎn)C作CE的垂線交BD于點(diǎn)G,設(shè)BD與AC的交點(diǎn)為Q.由tan∠BAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/0.png)
,得到
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/1.png)
.證明△BCG∽△ACE,得到
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/2.png)
.得到GB=DE,得到F是EG中點(diǎn).于是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/3.png)
,即可得到BE-DE=EG=2CF;
(3)分類討論:當(dāng)AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/4.png)
時(shí),取AB的中點(diǎn)M,連接MF和CM,tan∠BAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/5.png)
,且BC=6,計(jì)算出AC=12,AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/6.png)
.M為AB中點(diǎn),則CM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/7.png)
,F(xiàn)M=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/8.png)
=2.當(dāng)且僅當(dāng)M、F、C三點(diǎn)共線且M在線段CF上時(shí)CF最大,此時(shí)CF=CM+FM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/9.png)
;當(dāng)AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/10.png)
時(shí),取AB的中點(diǎn)M,連接MF和CM,類似于情況1,可知CF的最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/11.png)
.即可得到線段CF長度的最大值.
解答:解:(1)∵F為BD中點(diǎn),DE⊥AB,
∴CF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/12.png)
BD,EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/13.png)
BD,
∴CF=EF,
∴k=1;
故答案為1.
(2)如圖,過點(diǎn)C作CE的垂線交BD于點(diǎn)G,設(shè)BD與AC的交點(diǎn)為Q.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/images14.png)
由題意,tan∠BAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/14.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/15.png)
.
∵D、E、B三點(diǎn)共線,
∴AE⊥DB.
∵∠BQC=∠AQD,∠ACB=90°,
∴∠QBC=∠EAQ.
∵∠ECA+∠ACG=90°,∠BCG+∠ACG=90°,
∴∠ECA=∠BCG.
∴△BCG∽△ACE.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/16.png)
∴GB=DE.
∵F是BD中點(diǎn),
∴F是EG中點(diǎn).
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/images18.png)
在Rt△ECG中,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/17.png)
,
∴BE-DE=EG=2CF;
(3)情況1:如圖,當(dāng)AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/18.png)
時(shí),取AB的中點(diǎn)M,連接MF和CM,
∵∠ACB=90°,tan∠BAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/19.png)
,且BC=6,
∴AC=12,AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/20.png)
.
∵M(jìn)為AB中點(diǎn),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/images23.png)
∴CM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/21.png)
,
∵AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/22.png)
,
∴AD=4.∵M(jìn)為AB中點(diǎn),F(xiàn)為BD中點(diǎn),
∴FM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/23.png)
=2.
如圖:∴當(dāng)且僅當(dāng)M、F、C三點(diǎn)共線且M在線段CF上時(shí)CF最大,
此時(shí)CF=CM+FM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/24.png)
.
情況2:如圖,當(dāng)AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/25.png)
時(shí),取AB的中點(diǎn)M,連接MF和CM,
類似于情況1,可知CF的最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/26.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/images30.png)
綜合情況1與情況2,可知當(dāng)點(diǎn)D在靠近點(diǎn)C的
三等分點(diǎn)時(shí),線段CF的長度取得最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190823730441790/SYS201311011908237304417023_DA/27.png)
.
點(diǎn)評(píng):本題考查了三角形相似的判定與性質(zhì).也考查了旋轉(zhuǎn)的性質(zhì)和三角函數(shù)的定義以及直角三角形斜邊上的中線等于斜邊的一半.