Proof that a circle can be circumscribed around a quadrilateral

Question

I'm solving a problem you can see above: $PQRS$ is a trapezoid, $HM$ and $KN$ are legs bisectors. I'm supposed to prove that $\alpha = \beta$, which as you look closely can be reduced to proving $$\measuredangle SPM = \measuredangle NQR$$ (equivalently $\measuredangle PSM = \measuredangle NRQ$).

Hypothesis

These angles somehow resemble inscribed angles on arc $NM$. A circle could be circumsribed around $PQMN$ or $NMRS$. But I'm struggling with proving that one can actually circumsribe a circle around either of these. I'd appreciate some help from you.

Here you can find a brilliant drawing by User Raffaele who's done a great deal of work on the problem. However, a rigorous solution has not been provided by anyone.

Answer 1

This is not true which can be seen if it happens to be a non-rectangual parallelogram

We see that if $PQMN$ is to be circumscribed the center $C$ must form isosceles triangles $QMC$ and $PCN$ which means that $C$ must lie on the bisectioning normal to $QM$ and $PN$ which is not possible.

Answer 2

This is the solution I gave a month or so, but Jyrki Lahtonen deleted question because it appeared on some contest.

Here is the original question:

Let AB and CD be the bases of trapezoid $ABCD$. Bisectors of the arms AD , BC intersect the BC and AD segments respectively at points P and Q. Prove that $∠APD = ∠BQC $.

and here is my answer:

Say $M$ and $N$ are midpoints of $BC$ and $AD$ respectively.

Then $MNQP$ is cyclic since $\angle QNP = \angle QMP = \pi/2$.

Now let $\angle PNM = x$. Then $\angle MQP =x$ and then $$\angle PQM = \pi/2-x\;\;\;\;...(1)$$ Since $MN$ is middle line it is parallel to $DC$ and since $\angle QNM = \pi/2+x$ so is $$\angle QDC = \pi/2+x\;\;\;\;... (2)$$

From (1) and (2) we deduce that $QPCD$ is cyclic and thus conclusion.