Let $ABCD$ be a rectangle, $E$ midpoint of $\overline{DC}$ and $G$ point on $\overline{AC}$ such that $\vec{BG}$ is perpendicular to $\vec{AC}$. Also, let $F$ be a midpoint of $\overline{AG}$. Prove that angle $\angle BFE =\pi/2$.
I found already 4 solution which I have posted here Proving right angle using vectors
But now I'm interested in geometric transformation solution. I found this one also:
Spiral similarity $\mathcal{P}$ with center at $B$ that takes $A\mapsto D$ takes also $G\mapsto C$ and there for it takes midpoint $F$ of $AG$ to a midpoint of $CD$, that is $E$. Now this spiral similarity $\mathcal{P}$ induces new spiral similarity $\mathcal{P}'$ which takes $A\mapsto F$ and $D\mapsto E$ and also $B\mapsto B$, so we have $$\angle EFB = \angle DAB = 90^{\circ}$$
Finaly, here is my question. We notice also $$A\stackrel{\mathcal{P}}{\longmapsto}D \stackrel{\mathcal{S}}{\longmapsto}C \stackrel{\mathcal{P}^{-1}}{\longmapsto}G$$ where $\mathcal{S}$ is reflection across $E$. So transformation $$ \mathcal{T} = \mathcal{P}^{-1}\circ \mathcal{S} \circ \mathcal{P}$$ is reflection across $F$ that takes $A$ to $G$. Can we from here deduce that $\angle EFB = 90^{\circ}$?