【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=
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,得到
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.證明△BCG∽△ACE,得到
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.得到GB=DE,得到F是EG中點(diǎn).于是
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,即可得到BE-DE=EG=2CF;
(3)分類討論:當(dāng)AD=
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時(shí),取AB的中點(diǎn)M,連接MF和CM,tan∠BAC=
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,且BC=6,計(jì)算出AC=12,AB=
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.M為AB中點(diǎn),則CM=
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,F(xiàn)M=
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=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=
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;當(dāng)AD=
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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
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.即可得到線段CF長(zhǎng)度的最大值.
解答:解:(1)∵F為BD中點(diǎn),DE⊥AB,
∴CF=
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BD,EF=
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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.
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由題意,tan∠BAC=
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,
∴
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.
∵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.
∴
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∴GB=DE.
∵F是BD中點(diǎn),
∴F是EG中點(diǎn).
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在Rt△ECG中,
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,
∴BE-DE=EG=2CF;
(3)情況1:如圖,當(dāng)AD=
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時(shí),取AB的中點(diǎn)M,連接MF和CM,
∵∠ACB=90°,tan∠BAC=
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,且BC=6,
∴AC=12,AB=
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.
∵M(jìn)為AB中點(diǎn),
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∴CM=
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,
∵AD=
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,
∴AD=4.∵M(jìn)為AB中點(diǎn),F(xiàn)為BD中點(diǎn),
∴FM=
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=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=
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.
情況2:如圖,當(dāng)AD=
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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
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.
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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(zhǎng)度取得最大值為
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.
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.