Arc degree in the circumference

Question

In this image, $MN \parallel AT$, $CD=DA$ and $\angle PDT=90º$ then arc $NP$=?

I don't know how to apply the angle theorems inside this circle. I tried calling $\alpha$, $\beta$ and $\gamma$ the arcs $CM$, $MB$ and $PT$ respectively, and then applying equation systems, but it didn't work.

Any hints?

Answer 1

Notice that $DT$ is pependicular bisector for segment $AC$.

Since $$\angle NMT = \angle MTA = \angle MTA$$ we see that chord $CM$ and $NT$ are equal. So $CMTN$ is trapezoid and so $CN||MT$ and thus $PC\bot CN$, so $\angle NC P = 90^{\circ}$