For a given uniform square lamina $ABCD$ with centre $O$, let $I_{AC}$ and $I_{EF}$ be the moments of inertia about the diagonal $AC$ and the line $EF$ passing through $O$ and parallel to side $AB$ respectively. Then:

Let $I_x$ and $I_y$ be the moments of inertia about two mutually perpendicular axes in the plane of the lamina passing through the center.By the perpendicular axis theorem, the moment of inertia about the axis perpendicular to the plane ($I_z$) is $I_z=I_x+I_y$.For a square, due to symmetry, any two perpendicular axes in the plane passing through the center are equivalent.For axes parallel to the sides ($EF$ and its perpendicular): $I_z=I_{EF}+I_{EF}=2I_{EF}$.For the diagonals ($AC$ and $BD$): $I_z=I_{AC}+I_{BD}=2I_{AC}$.Equating the two expressions for $I_z$:
$$2I_{AC}=2I_{EF}⟹I_{AC}=I_{EF}$$