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