Let the each side of square lamina is $d$.
So, $ I_{E F}=I_{G H}$ (due to symmetry)
and $ I_{A C}=I_{B D}$ (due to symmetry)
and $ I_{A C}=I_{B D} $ (due to symmetry)
Now, according to theorem of perpendicular axis,
$ I_{A C}+I_{B D}=I_{0}$
$\Rightarrow 2 I_{A C}=I_{0} $
and $ I_{E F}+I_{G H}=I_{0}$
$\Rightarrow 2 I_{E F}=I_{0}$
From Eqs. (i) and (ii), we get $I_{A C}=I_{E F}$
$\therefore I_{A D} =I_{E F}+\frac{m d^{2}}{4} $
$=\frac{m d^{2}}{12}+\frac{m d^{2}}{4}\left(\text { as } I_{E F}=\frac{m d^{2}}{12}\right)
$
So, $ I_{A D}=\frac{m d^{2}}{3}=4 I_{E F}$
