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解:(1)∵四邊形ABCD是菱形,且菱形ABCD的邊長(zhǎng)為2cm,
∴AB=BC=2,∠BAC=
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∠DAB,
又∵∠DAB=60°(已知),
∴∠BAC=∠BCA=30°;
如圖1,連接BD交AC于O.
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∵四邊形ABCD是菱形,
∴AC⊥BD,OA=
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AC,
∴OB=
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AB=1(30°角所對(duì)的直角邊是斜邊的一半),
∴OA=
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(cm),AC=2OA=2
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(cm),
運(yùn)動(dòng)ts后,
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,
∴
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又∵∠PAQ=∠CAB,
∴△PAQ∽△CAB,
∴∠APQ=∠ACB(相似三角形的對(duì)應(yīng)角相等),
∴PQ∥BC(同位角相等,兩直線平行)
(2)如圖2,⊙P與BC切于點(diǎn)M,連接PM,則PM⊥BC.
在Rt△CPM中,∵∠PCM=30°,∴PM=
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PC=
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由PM=PQ=AQ=t,即
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=t
解得t=4
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-6,此時(shí)⊙P與邊BC有一個(gè)公共點(diǎn);
如圖3,⊙P過點(diǎn)B,此時(shí)PQ=PB,
∵∠PQB=∠PAQ+∠APQ=60°
∴△PQB為等邊三角形,∴QB=PQ=AQ=t,∴t=1
∴
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時(shí),⊙P與邊BC有2個(gè)公共點(diǎn).
如圖4,⊙P過點(diǎn)C,此時(shí)PC=PQ,即2
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t=t,∴t=3-
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.
∴當(dāng)1<t≤3-
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時(shí),⊙P與邊BC有一個(gè)公共點(diǎn),
當(dāng)點(diǎn)P運(yùn)動(dòng)到點(diǎn)C,即t=2時(shí),⊙P過點(diǎn)B,此時(shí),⊙P與邊BC有一個(gè)公共點(diǎn),
∴當(dāng)t=4
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-6或1<t≤3-
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或t=2時(shí),⊙P與菱形ABCD的邊BC有1個(gè)公共點(diǎn);
當(dāng)4
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-6<t≤1時(shí),⊙P與邊BC有2個(gè)公共點(diǎn).
分析:(1)連接BD交AC于O,構(gòu)建直角三角形AOB.利用菱形的對(duì)角線互相垂直、對(duì)角線平分對(duì)角、鄰邊相等的性質(zhì)推知△PAQ∽△CAB;然后根據(jù)“相似三角形的對(duì)應(yīng)角相等”證得∠APQ=∠ACB;最后根據(jù)平行線的判定定理“同位角相等,兩直線平行”可以證得結(jié)論;
(2)如圖2,⊙P與BC切于點(diǎn)M,連接PM,構(gòu)建Rt△CPM,在Rt△CPM利用特殊角的三角函數(shù)值求得PM=
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PC=
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,然后根據(jù)PM=PQ=AQ=t列出關(guān)于t的方程,通過解方程即可求得t的值;
如圖3,⊙P過點(diǎn)B,此時(shí)PQ=PB,根據(jù)等邊三角形的判定可以推知△PQB為等邊三角形,然后由等邊三角形的性質(zhì)以及(2)中求得t的值來確定此時(shí)t的取值范圍;
如圖4,⊙P過點(diǎn)C,此時(shí)PC=PQ,據(jù)此等量關(guān)系列出關(guān)于t的方程,通過解方程求得t的值.
點(diǎn)評(píng):本題綜合考查了菱形的性質(zhì)、直線與圓的位置關(guān)系以及相似三角形的判定等性質(zhì).解答(2)題時(shí),根據(jù)⊙P的運(yùn)動(dòng)過程來確定t的值,以防漏解.