how to show that the angle contained by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining 2 angles?

Answer :

From given information , we form our diagram , As  :

Here  $\angle$ BAO  = $\angle$ OAD  =  $\frac{\angle \hspace{0.17em}BAD}{2}$                                      
And
$\angle$ABO  = $\angle$ OBC =  $\frac{\angle \hspace{0.17em}ABC}{2}$
And
To prove :  $\angle$ AOB  =  $\frac{\angle \hspace{0.17em}BCD + \angle CDA}{2}$ 

So,

We know from angle sum property of quadrilateral , So , for ABCD , we get

$\angle$ BAD  +  $\angle$ ABC  + $\angle$ BCD +  $\angle$ CDA  =  360$°$                                                                     -------------- ( 1 )

And

We know from angle sum property of triangle , So , for AOB , we get

$\angle$ BAO  +  $\angle$ ABO  + $\angle$ AOB  =  180$°$    , Substitute given value , we get 

  $\frac{\angle \hspace{0.17em}BAD}{2}$   + $\frac{\angle \hspace{0.17em}ABC}{2}$ + $\angle$ AOB  =  180$°$  , Taking LCM we get

$\angle$ BAD  +  $\angle$ ABC + 2 $\angle$ AOB  =  360$°$                                                                                     -------------- ( 2 )

Now we subtract equation 2 from equation 1 , we get

$\angle$ BCD +  $\angle$ CDA  - 2 $\angle$ AOB =   0

 2 $\angle$ AOB =  $\angle$ BCD +  $\angle$ CDA 

$\angle$ AOB  = $\frac{\angle \hspace{0.17em}BCD + \angle CDA}{2}$                                                                           ( Hence proved )