The diagonals of the parallelogram whose sides are $lx+my+n=0,lx+my+n_1=0,mx+ly+n=0,mx+ly+n_1=0$  include an angle

$$\begin{gathered} lx+my+n=0\to \left(1\right) ,lx+my+n_1=0\to \left(2\right), \\ mx+ly+n=0\to \left(3\right),mx+ly+n_1=0\to \left(4\right) \end{gathered}$$

$$\begin{gathered} lines \left(1\right),\left(2\right) are parallel and \left(3\right),\left(4\right) are parallel \\ and dis\tan ce between parallel lines \left(1\right),\left(2\right) and \left(3\right),\left(4\right) are equal \left(=\left|\frac{n-n_1}{\sqrt{l^2+m^2}}\right|\right) \end{gathered}$$

Hence the given four lines form a Rhombus.

$\therefore$  Angle between diagonals $=\frac{\pi }{2}$