Let $D$ be the point on the base $BC$ of an isosceles triangle $ABC$ with $AB=AC$ and $m(BAC)=80^\circ$, such that $BD : DC = 2 : 1$, and let $P$ be the point on the segment $AD$ such that $m(BAC) = m(BPD)$. Find $m(DPC)$ in degrees.

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