Let the each side of square lamina is $d$.
So, $ I_{E F}=I_{G H}$ (due to symmetry)
and $I_{A C}=I_{D B}$ (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 $(i)
$\text { and } I_{E F}+I_{G H}=I_0$
$\Rightarrow 2 I_{E F}=I_0$(ii)
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}$
