A challenging straightedge and compass construction

Thanks to Xaver, a simple solution (but a not-so-trivial one).

Lemma 1. If $P\in OA$ and $Q\in OB$ fulfills $PA=QB$, then $PB\cap QA$ lies on a line
that is parallel to the angle bisector of $\widehat{AOB}$.

Lemma 2. $C=AB'\cap A'B$ lies on a fixed circle $\Gamma$ through the incenter of $AOB$,
since $\widehat{ACB}=\frac{\pi+\theta}{2}$.

So, let $P$ be a point of $OA$ and $Q$ be the corresponding $Q$-point on $OB$, as in Lemma 1.

Let $D=BP\cap AQ$ and $\ell$ be the line through $D$ that is parallel to the angle bisector of $\widehat{AOB}$.

Let $I$ be the incenter of $AOB$ and $\Gamma$ the circumcircle of $AIB$.

Then $\color{blue}{C=\ell \cap \Gamma}$ and $A',B'$ are easily found.

Let $S$ be the intersection of $AB'$ and $BA'$. Let $\varphi:=\angle ASA'$. Then $\theta+2\varphi=180°$ always holds.

I haven't really worked out the details, but I'm sure that this helps to construct the segments $AB'$ and $BA'$ (and therefore $A'$ and $B'$).