There are two well-known results in cyclic quadrilaterals:
(a) Opposite angles add upto $180^\circ$.
(b) Angles in the same segments are equal.
The converse of these two results are also true. The proof of (b) is quite standard and (a) can be proved by using (b) which is also well-known.
Here my concern starts:
I want to prove (b) from (a) directly, means I want to prove (a) $\Rightarrow$ (b). One (easy) way of doing this:
(a) $\Rightarrow$ quadirateral is cyclic $\Rightarrow$ (b). (In this method I used the converse of (a)).
But I want to remove the intermediate phase ('quadrilateral is cyclic'), I want to go directly to (b) from (a).
Here is my attempt:
Since, we are given: \begin{align} y_1+y_2+w_1+w_2&=180^\circ\\ x_1+x_2+z_1+z_2&=180^\circ. \end{align} We want to show (one possibility): $$ y_1=z_1. $$
By using some geometry we get the following relations: \begin{align} y_2+z_2&=w_1+x_2\\ x_1+y_1&=z_1+w_2\\ w_1+y_1&=z_1+z_2\\ x_1+z_2&=w_1+w_2\\ y_2+w_2&=x_1+x_2\\ z_1+x_2&=y_1+y_2. \end{align}
I tried doing elimination to get $y_1=z_1$ (or something equivalent) but could not get it. Please help.