Let $ABC$ be any triangle . Two squares $BAEP$ and $CAFR$ are constructed externally to $ABC$.Let $M$ be the midpoint og $BC$. Show that $AM$ is orthogonal to $EF$.
I have no idea how to prove its perpendicular.
Maybe both triangles $ABC$ and $EAF$ share this property.
Thanks!
Since the question is posed without much context, here's a solution using vectors. (I myself would prefer an Euclidean-geometrical approach.)
Since $$\begin{aligned}\vec{AM}&=\frac12(\vec{AB}+\vec{AC}),\\ \vec{EF}&=\vec{AF}-\vec{AE}\end{aligned}$$ it suffices to show that $$(\vec{AB}+\vec{AC})(\vec{AF}-\vec{AE})=0.\tag{1}$$ But the LHS of (1) is $$ \vec{AB}\cdot \vec{AF}+\vec{AC}\cdot \vec{AF}-\vec{AB}\cdot \vec{AE}-\vec{AC}\cdot\vec{AE}.$$ Since $AC\perp AF$ and $AB\perp AE$, we only need to show $$\vec{AB}\cdot\vec{AF}=\vec{AC}\cdot\vec{AE}.\tag{2}$$ But (2) is clear since $AB=AE$, $AC=AF$, and $\angle BAF=\angle EAC$.
Euclidean style:
Let $A^\prime$ be the reflection of $A$ about $M$, so $AC\parallel BA^\prime$. It follows that $$\angle ABA^\prime=\angle EAF=180^\circ-\angle BAC.$$ Thus, $\triangle AEF=\triangle ABA^\prime$. So $\angle AEF=\angle BAA^\prime$. Simple angle chasing should imply that $AA^\prime\perp EF$.
Note that we proved even more that $EF=2AM$.