Four touching circles and one common tangent 
Given four touching circles and one common tangent, show that $$\angle BAD = \dfrac{1}{2}(\angle DO_1A + \angle AO_2B)$$


It is done by looking at triangles $\triangle ADO_1$ and $\triangle ABO_2$. 
Now I am trying to prove that a circle can be circumscribed around $ABCD$. 
I tried to show $\angle ADC + \angle DCB = 180 ^\circ$, thus $AD || CB$ but I messed up (my idea was to prove that $ABCD$ is an isosceles trapezium). 
My second thought was to show $\angle DCB + \angle DAB = 180 ^\circ$, but I don't see how this can be done.
Would appreciate help of any kind.

It will be easy to prove sum of opposite angles of quadrilateral ABCD is 180. Hence it is cyclic quadrilateral.
