已知,點D是等邊△ABC內的任一點,連接OA,OB,OC. (1)如圖1,己知∠AOB=150°,∠...
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問題詳情:
已知,點D是等邊△ABC內的任一點,連接OA,OB,OC.
(1)如圖1,己知∠AOB=150°,∠BOC=120°,將△BOC繞點C按順時針方向旋轉60°得△ADC.
①∠DAO的度數是_______________
②用等式表示線段OA,OB,OC之間的數量關係,並*;
(2)設∠AOB=α,∠BOC=β.
①當α,β滿足什麼關係時,OA+OB+OC有最小值?請在圖2中畫出符合條件的圖形,並説明理由;
②若等邊△ABC的邊長為1,直接寫出OA+OB+OC的最小值.
【回答】
解:(1)①90°. ······························································································ 2分
②線段OA,OB,OC之間的數量關係是. ····························· 3分
如圖1,連接OD.···························································································· 4分
∵△BOC繞點C按順時針方向旋轉60°得△ADC,
∴△ADC≌△BOC,∠OCD=60°.
∴CD = OC,∠ADC =∠BOC=120°,AD= OB.
∴△OCD是等邊三角形,················································································ 5分
∴OC=OD=CD,∠COD=∠CDO=60°,
∵∠AOB=150°,∠BOC=120°,
∴∠AOC=90°,
∴∠AOD=30°,∠ADO=60°.
∴∠DAO=90°.································································································· 6分
在Rt△ADO中,∠DAO=90°,
∴.
∴.·························································································· 7分
(2)①如圖2,當α=β=120°時,OA+OB+OC有最小值. ····································· 8分
作圖如圖2,··································································································· 9分
如圖2,將△AOC繞點C按順時針方向旋轉60°得△A’O’C,連接OO’.
∴△A′O′C≌△AOC,∠OCO′=∠ACA′=60°.
∴O′C= OC, O′A′ = OA,A′C = BC,
∠A′O′C =∠AOC.
∴△OC O′是等邊三角形.··············································································· 10分
∴OC= O′C = OO′,∠COO′=∠CO′O=60°.
∵∠AOB=∠BOC=120°,
∴∠AOC =∠A′O′C=120°.
∴∠BOO′=∠OO′A′=180°.
∴四點B,O,O′,A′共線.
∴OA+OB+OC= O′A′ +OB+OO′ =BA′ 時值最小.············································ 11分
②當等邊△ABC的邊長為1時,OA+OB+OC的最小值A′B=. ··················· 12分
知識點:等腰三角形
題型:解答題
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