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，並說明理由；
②若等邊△ABC的邊長爲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的邊長爲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
