since ax^2 +2hxy +by^2 = 0 is homogenous eqn second degree we can take two lines as y=px ; y=rx
lx +my=1
pr=a/b
p+r=-2h/b
solve to get vertices of triangle
(0,0)
(1/l+mp , p/l+mp)
(1/l+mr , r/l+mr)
alpha=[2l +m(p+r)]/[3*(l+mp)(l+mr)]
alpha=2*[bl - hm] /[3*(l+mp)(l+mr)]
alpha/bl-hm =2/[3*(l+mp)(l+mr)]
similarly
beta/am-hl =2/[3*(l+mp)(l+mr)]
RHS=2/[3*[l^2+lm(p+r) +m^2.pr]
=2/3*[b.l^2 -2hlm +am^2]
hence proved