Given \(\triangle ABC\), let \(D\) be a point on \(AB\) produced beyond \(B\), such that \(BD=BC\), and let \(E\) be a point on \(AC\) produced beyond \(C\), such that \(CE=BC\). Let \(P\) be the intersection of \(BE\) and \(CD\), and suppose that:

\[\large{\dfrac{DP}{BE}+ \dfrac{EP}{CD} = 2\sin \left( \dfrac{\angle BAC}{2} \right) }\]

Submit the value of \(\angle BAC\) in degrees as your answer.

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