设三角形ABC不是直角三角形,O为三角形ABC的外心,H为垂心,直线OH交AC于K,交AB于L,若OK=HL,求角A的值

记三角形三边BC,AC,AB为a,b,c,外接圆半径为R
作OM⊥AC于M,BH⊥AC于N
则OM = RcosB
HN = CNcot∠CHN = acosCcotA
则KO/KH = OM/HN = RcosB/(acosCcotA) = cosB/(2sinAcosCcotA) = cosB/(2cosCcosA)
同理
LH/LO = 2cosAcosB/cosC)
由OK = HK有KO/KH = LH/LO
所以2cosA = 1/(2cosA)
cosA = ±1/2
A = π/3或2π/3