Dot product proof in a quadrilateral

Question

ABCD is a quadrilateral such that BC = AD. Let E be the midpoint of AB and F be the midpoint of CD. Show that: $$\vec {EF} \cdot \vec {BC} = \vec {EF} \cdot \vec {AD}$$ I suspect that it has something to do with projections but I'm not sure how to proceed from that idea. Any help would be appreciated.

Answer 1

Let the sides of the quadrilateral be $\vec a$, $\vec b$, $\vec c$ and $\vec d$. $$\vec{EF}\cdot \vec{BC}=(\frac{1}{2}\vec a+\vec b +\frac{1}{2}\vec c)\cdot\vec b=(-\frac{1}{2}\vec b+\vec b -\frac{1}{2}\vec d)\cdot\vec b=\frac{1}{2}(\vec b-\vec d)\cdot\vec b$$ $$\vec{EF} \cdot \vec{AD}=(-\frac{1}{2}\vec a-\vec d -\frac{1}{2}\vec c)\cdot-\vec d=(\frac{1}{2}\vec b-\vec d +\frac{1}{2}\vec d)\cdot-\vec d=-\frac{1}{2}(\vec b-\vec d)\cdot\vec d$$ Since $BC = AD$: $$|\vec b|^2=|\vec d|^2$$$$(\vec b-\vec d)\cdot(\vec b+\vec d)=0$$ Hence: $$(\vec b-\vec d)\cdot\vec b=-(\vec b-\vec d)\cdot\vec d$$Therefore: $$\vec{EF} \cdot \vec{BC}=\frac{1}{2}(\vec b-\vec d)\cdot\vec b=-\frac{1}{2}(\vec b-\vec d)\cdot\vec d=\vec{EF} \cdot \vec{AD}$$