A point $P$ is marked within regular hexagon $ABCDEF$ so that we have $angle BAP=angle DCP = 50^circ $ If $angle APB$ has measure $x$ degrees, find $x$.

Here's a diagram:

Referring to the diagram, I've tried to do some angle chasing:

$angle PAF=angle PGE=angle CPH=angle PCB=70, angle PGF=110, angle GPA=HPC=60$, and all the interior angles are equal to $120$.

However, it seems like this isn't sufficient to solve the problem and I think I might need a construction of some sort - but I'm stuck here.

Assuming the solution needs more than just angle chasing, I'd appreciate constructions/observations and explanations of why they'd solve the problem.

Solution

As the figure shows, we may readily obtain that $$angle OAP=angle OCP=10^o,$$ which implies that $A,O,P,C$ are concyclic. But we know the center of the circle $(AOC)$ is $B$, hence $$BA=BP.$$
As a result, $$angle APB=angle BAP=50^o.$$

Suppose we construct an alternative point $P'$ in a slightly different way. Draw the line through $C$ of the hexagon such that $angle DCG = 50$, same as in your diagram, and ignore all the rest. Then draw point $P'$ on that line such that $BC=BP'$. That gives this picture:

Using the fact that $AB=BC=BP'$ you now have isosceles triangles $BCP'$ and $BP'A$, and you should have no trouble finding all the angles in those two triangles.

Happily it turns out that $angle BAP' = 50$, so that the point $P'$ is actually the same as point $P$, and $angle APB = angle AP'B$.