$I_1,I_2,I_3,I_4$ are respectively the moments of inertia of a thin square plate ABCD of uniform thickness about axes 1,2,3 and 4 which are in the plane of the plate. The moment of inertia about an axis passing through the centre O and perpendicular to the plane of the plate is

Because of symmetry of axes 1 and 2, $I_1=I_2\text{.}$ Similarly, $I_3=I_4$ From perpendicular axes theorem, it follows that the moment of inertia of the plate about an axis passing through the centre and perpendicular to the plane of the plate is 

$I=I_1+I_2=I_3+I_4=2I_1=2I_3\left(\because I_1=I_2,I_3=I_4\right)\text{ or }{I}_1=I_3$ 

Thus, $I=I_1+I_2=I_3+I_4=I_1+I_3$