∴sin2B=sinA•sinC.
设l1:a1x+b1y+c1=0,l2:a2x+b2y+c2=0.
∵
a1 |
a2 |
sin2A |
sin2B |
sin2A |
sinAsinC |
sinA |
sinC |
b1 |
b2 |
sinA |
sinC |
c1 |
c2 |
−a |
−c |
−2RsinA |
−2RsinC |
sinA |
sinC |
∴
a1 |
a2 |
b1 |
b2 |
c1 |
c2 |
∴l1与l2重合,
故答案为重合.
a1 |
a2 |
sin2A |
sin2B |
sin2A |
sinAsinC |
sinA |
sinC |
b1 |
b2 |
sinA |
sinC |
c1 |
c2 |
−a |
−c |
−2RsinA |
−2RsinC |
sinA |
sinC |
a1 |
a2 |
b1 |
b2 |
c1 |
c2 |