Four touching circles and one common tangent

Question

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.

Answer 1

Hint: As you proved $$<BAD = {1\over 2}(<AO_2B+ <AO_1D)$$

similary we have also: $$<BCD = {1\over 2}(<CO_3B+ <CO_4D)$$

so $$<BAD +<BCD = {1\over 2}(<AO_2B+ <AO_1D)+{1\over 2}(<CO_3B+ <CO_4D)$$ $$={1\over 2}360 = 180$$