\({I_z} = {I_x} + {I_y}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,or,\,\,{I_z} = 2{I_y}\)

\(\begin{array}{l}
\left( {\,{I_x} = {I_y}\,by\,symmetry\,of\,the\,figure} \right)\\
\therefore \,{I_{EF}} = \frac{{{I_z}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
\end{array}\)

Again, by the same theorem

\(\begin{array}{l}
{I_z} = {I_{AC}} + {I_{BD}} = 2{I_{AC}}\\
\left( {\therefore \,{I_{AC}} = {I_{BD}}\,by\,symmetry\,of\,the\,figure} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,{I_{AC}} = \frac{{{I_z}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)
\end{array}\)

    From \((i)\) and \((ii)\), we get \({I_{EF}} = {I_{AC}}\)