Sketch: Assume the contrary and after that consider a triangle of a minimum area,denote it as $ \phi$.The other three points must be outside the $ \phi(DEF)$.Denote the trianlge formed by other three points as $ \gamma(ABC)$.Now consider two cases: If $ \gamma$ contains $ \phi$ and if it doesn't.First case: suppose that ray $ FE$ intersects $ BC$ at some point.Then clearly $ \triangle BFC$ contains $ E$ but it does not contain $ A$.Contradiction. Second case:similar approach works here...some boring work with considering many cases.

We can consider the convex closure and work in cases. It's not hard to imagine that.