Let $D$ be the point on the base $BC$ of an isosceles $\vartriangle ABC$ triangle such that $\frac{\left| BD \right|}{\left| DC \right|}=\text{ }2$, and let $P$ be the point on the segment $\left[ AD \right]$ such that $\angle BAC=\angle BPD$. Prove that $\angle DPC=\frac{1}{2}\angle BAC$.