if ax^3+by^3+cx^2y+dxy^2=0 represent three distinct straight lines,such that each line bisects the angle between other two then prove that 3b+c=0

Let the lines be

y-m_ix = 0 where i=1,2,3

As one line is equally inclined to the other two, we impose the following condition.

(m₁-m₃)/(1+m₁m₃) = -(m₂-m₃)/(1+m₂m₃)

Simplify to get,

(m₁+m₂+m₃)-3m₃-m₃(m₁m₂+m₂m₃+m₃m₁)+3m₁m₂m₃ = 0       ...........(i)

Similarly applying the condition on the other two pairs, we get,

(m₁+m₂+m₃)-3m₂-m₂(m₁m₂+m₂m₃+m₃m₁)+3m₁m₂m₃ = 0         ............(ii)

and

(m₁+m₂+m₃)-3m₁-m₁(m₁m₂+m₂m₃+m₃m₁)+3m₁m₂m₃ = 0          .......... (iii)

Adding equations (i),(ii) and (iii), we get,

-(m₁m₂+m₂m₃+m₃m₁)(m₁+m₂+m₃) + 9m₁m₂m₃ = 0                 ............(iv)

Putting y=mx in the equation of combined equation,

we get,

bm³ + dm² + cm + a = 0

By theory of equations,

m₁+m₂+m₃ = -d/b

m₁m₂+m₂m₃+m₃m₁ = c/b

m₁m₂m₃ = -a/b

Putting the above relations in equation (iv),

-(c/b)(-d/b) + 9(-a/b) = 0

cd-9ab=0

This is the condition that i get. No idea where i went wrong.