How to find that o is the midpoint of BD
 

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{It} \mathrm{depends} \mathrm{upon} \mathrm{the} \mathrm{quadrilateral} \mathrm{if} \mathrm{it} \mathrm{is} \mathrm{parallelogram}, \mathrm{or} \mathrm{rhombus} \mathrm{or} \mathrm{kite} \\ \mathrm{as} \mathrm{it} \mathrm{appears} \mathrm{it} \mathrm{is} \mathrm{kite} \mathrm{then} \mathrm{O} \mathrm{will} \mathrm{be} \mathrm{mid} \mathrm{point}. \mathrm{It} \mathrm{is} \mathrm{property} \mathrm{of} \mathrm{kite}. \end{gathered}$$
And be more precise O will be mid point of BD only because then for this Kite, 
BC  = CD , again property of kite 
and 
OC is common and angle BOC = angle COD = 90 
RHS congruency
so triangle COB and triangle COD will be congruent so 
BO = OD