已知，点D是等边△ABC内的任一点，连接OA，OB，OC.   (1)如图1，己知∠AOB＝150°，∠...

问题详情：

    已知，点D是等边△ABC内的任一点，连接OA，OB，OC.

    (1)如图1，己知∠AOB＝150°，∠BOC＝120°，将△BOC绕点C按顺时针方向旋转60°得△ADC．

①∠DAO的度数是_______________

    ②用等式表示线段OA，OB，OC之间的数量关系，并*；

    (2)设∠AOB＝α，∠BOC＝β.

    ①当α，β满足什么关系时，OA＋OB＋OC有最小值？请在图2中画出符合条件的图形，并说明理由；

    ②若等边△ABC的边长为1，直接写出OA＋OB＋OC的最小值．

【回答】

解：（1）①90°. ······························································································ 2分

②线段OA，OB，OC之间的数量关系是.  ····························· 3分

如图1，连接OD.···························································································· 4分

∵△BOC绕点C按顺时针方向旋转60°得△ADC，

∴△ADC≌△BOC，∠OCD=60°.

∴CD = OC，∠ADC =∠BOC=120°，AD= OB.

∴△OCD是等边三角形，················································································ 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如图2，当α=β=120°时，OA+OB+OC有最小值. ····································· 8分

作图如图2，··································································································· 9分

如图2，将△AOC绕点C按顺时针方向旋转60°得△A’O’C，连接OO’.

∴△A′O′C≌△AOC，∠OCO′=∠ACA′=60°.

∴O′C= OC, O′A′ = OA，A′C = BC,

∠A′O′C =∠AOC.

∴△OC O′是等边三角形.··············································································· 10分

∴OC= O′C = OO′，∠COO′=∠CO′O=60°.

∵∠AOB=∠BOC=120°，

∴∠AOC =∠A′O′C=120°.

∴∠BOO′=∠OO′A′=180°.

∴四点B，O，O′，A′共线.

∴OA+OB+OC= O′A′ +OB+OO′ =BA′ 时值最小.············································ 11分

②当等边△ABC的边长为1时，OA+OB+OC的最小值A′B=. ··················· 12分

知识点：等腰三角形

题型：解答题