==> $\frac{{{K_1}A({\theta _A} - {\theta _C})}}{l} = \frac{{{K_2}A({\theta _C} - {\theta _B})}}{l}$

==> $\frac{{{\theta _A} - {\theta _C}}}{{{\theta _C} - {\theta _B}}} = \frac{{{K_2}}}{{{K_1}}}$...$(i)$

Also ${\left( {\frac{Q}{t}} \right)_{AD}} = {\left( {\frac{Q}{t}} \right)_{DB}}$

==>$\frac{{{K_3}A({\theta _A} - {\theta _D})}}{l} = \frac{{{K_4}A({\theta _D} - {\theta _B})}}{l}$

==>$\frac{{{\theta _A} - {\theta _D}}}{{{\theta _D} - {\theta _B}}} = \frac{{{K_4}}}{{{K_3}}}$ ...$(ii)$

It is given that ${\theta _C} = {\theta _D},$ hence from equation $(i)$ and $(ii)$ we get $\frac{{{K_2}}}{{{K_1}}} = \frac{{{K_4}}}{{{K_3}}}$ 

==> ${K_1}{K_4} = {K_2}{K_3}$