\(\triangle ABC\) satisfies \(AB=AC\), \(\angle BAC=36^{\circ}\) and has circumradius \(2\). Let \(D\) be a point on the circumcircle \(ABC\) such that \(OD||BC\), where \(O\) is the circumcenter. Project \(D\) onto \(AB,AC\) to obtain \(E,F\), i.e. \(DE\perp AB, DF\perp AC\).

Suppose \(H\) is the orthcenter of \(ABC\). Then the area of \(\triangle HEF\) can be expressed as \(\sqrt {\frac {x+\sqrt {m}}{l}}\), where \(x,m,l\) are positive integers, and \(m\) is square-free. Find \(xml\)

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