Let $(h,k)$ be the intersection of the two given lines its equidistant from both axes.

$\therefore \left|h\right|=\left|k\right|$

Since it lies in the fourth quadrant, the point can be written as $(h,-h)$, where $h>0$.

$(h,-h)$ lies on both the lines.

$\therefore 4ah-2ah+c=0$

$\therefore c=-2ah$   $\therefore h=\frac{-c}{2a}$

$5bh-2bh+d=0$

$\therefore d=-3bh$   $\therefore h=\frac{-d}{3b}$

$\therefore \frac{-c}{2a}=\frac{-d}{3b}$

$\therefore 3bc-2ad=0$