What is the value of the semiperimeter of the quadrilateral $PQRS$?

Question

For reference: $ABCD$ is a parallelogram, $O$ it´s center and $MC = 8$. Calculate the quadrilateral $PQRS$ semiperimeter.

My progress:

Teorema Pitot: $SR + PQ = PS + QR$

Can I say $PS = RQ$? Can I say $MC = HA$?

Answer 1

Assuming $ \small \small AM$ is going through the center of the first circle as shown in the first diagram, then $ \small AM$ is angle bisector of $ \small \angle A$. You can show $ \small AB = BM$.

Then using $ \small BP + AS = AB + PS, PQ + SR = PS + QR$, $ \small QC + RD = QR + CD$,
show that $ \small AD + BC = AB + CD + 2 (PS + QR)$

You also have $ \small AB = CD = BM, AD = BC = BM + MC$

Can you take it from here?

Answer 2

We can get: $$P_{PQRS}=4BC-P_{ABCD}=2(BC-AB)=2MC=16.$$ Thus, $$\frac{1}{2}P_{PQRS}=8.$$