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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 $\angle BFE =\pi/2$.

So, I need to prove that $\vec{EF}\cdot \vec{FB}=0$. I tried to just express these vectors as sums of vectors of rectangle but haven't find any "elegant" way to prove it.

I did found one "ugly" way to prove it:

Let $|AB|=a$, $|AD|=b$ and $\vec{AB}=\vec{a}$, $\vec{AD}=\vec{b}$, so $\vec{AC}=\vec{a}+\vec{b}$ and since $\vec{a}\cdot\vec{b}=0$ we have $|AC|^2=a^2+b^2$. We can see that $\triangle ABC$ and $\triangle BCG$ are similar so $$|CG|=|CB|^2/|AC|$$ and since $\vec{CG}=\lambda \cdot \vec{CA}$ we get $$\lambda=\frac{b^2}{a^2+b^2}.$$

Now, we can express $\vec{EF}$ and $\vec{FB}$ in the terms of $\vec{a}$, $\vec{b}$, more precisely: $$\vec{EF}=\frac{1}{2}\left(\frac{-b^2}{a^2+b^2}\right)\vec{a}+\frac{1}{2}\left(\frac{a^2}{a^2+b^2}-2\right)\vec{b}$$ and $$\vec{FB}=\frac{1}{2}\left(2-\frac{a^2}{a^2+b^2}\right)\vec{a}+\frac{1}{2}\left(\frac{-a^2}{a^2+b^2}\right)\vec{b}$$ and if we multiply we get that dot product is $0$.

But as you can see, this $\lambda$ is "weird" and I am wondering if someone sees a more elegant way to prove this ?

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    $\begingroup$ Actually, if you write those final coefficients in terms of $\lambda$, it makes the proof of orthogonality much prettier. $\endgroup$ – J.G. Oct 9 '18 at 19:26
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You can simplify and generalize at the same time.

Instead of defining $E$ and $F$ as midpoints, define them as weighted averages:

$$ E=kC+(1-k)D $$ $$ F=kG+(1-k)A $$

(In your question $k=1/2.)$ As $k$ goes from $0$ to $1$, $\triangle{BFE}$ interpolates the similar triangles $\triangle{BAD}$ and $\triangle{BGC}$. And it is also similar to these two triangles.

Given that $(B-G)\cdot(C-G)=0$ and $(B-A)\cdot(D-A)=0$, you can show that $(B-F)\cdot(E-F)=0$ (the simplification is a bit tedious, and I used the fact that $(Rx)\cdot y+(Ry)\cdot x=0$ for perpendicular vectors $x,y$ and rotation $R$).

More about the phenomenon of linearly interpolating two directly similar figures can be found by Googling the terms 'spiral similarity' and 'fundamental theorem of directly similar figures'. Also, instead of using vectors you can use complex numbers, as shown in $\textbf{2.2}$ of Zachary Abel's Mean Geometry paper.

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Let $H$ be a midpoint of $BG$. Then $HF$ is a middle line in triangle $ABG$, so $FH||AB||CD$ and $FH = {1\over 2}AB = CE$ so $FHCE$ is a parallelogram, so $FE||CH$.

Now $HF\bot BC$ so $HF$ is (second) altitude in $\triangle BCF$ so $H$ is orthocenter in this triangle, since $BG$ is also altitude in this triangle. So line $CH\bot FB$ and thus $\angle BFE = 90^{\circ}$.

enter image description here

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Here is a vector solution:

Let $G$ be an origin of position vecotors. Then $G=0$, $F = A/2$, $D = A-B+C$, $A\cdot B = C\cdot B=0$ and $$E = {1\over 2}(C+D) = {1\over 2}(A-B+2C)$$

\begin{eqnarray} \vec{FB}\cdot \vec{FE} &=& (E-F)(B-F)\\ &=& {1\over 4}(-B+2C)(2B-A)\\ &=& -{1\over 2}(B^2+CA)\\ &=&0 \end{eqnarray}

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A solution with a help of the coordinate system.

Let $B = (0,0)$, $C= (2c,0)$ and $A= (0,2a)$. Then $D(2c,2a)$ and $E(2c,a)$. A perpendicular to $$AC:\;\;{x\over 2c}+{y\over 2a}=1$$ through $B$ is $$y={c\over a}x$$ which cuts $AC$ at $$G= \big({2a^2c\over a^2+c^2},{2c^2a\over a^2+c^2} \big)$$

so $$ F = \big({a^2c\over a^2+c^2},{2c^2a+a^3\over a^2+c^2} \big)$$

Now it is not difficult to see that $\vec{FE}\cdot \vec{FB} =0$.

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Another solution:

Say $M$ is midpoint of $AB$, then $FM$ is a middle line in triangle $ABG$ so $FM||BG$ so $\angle MFC = 90^{\circ}$ and thus $F$ is on a circle through $B,C$ and $M$. But on this circle is also $E$ since $\angle MEC = 90^{\circ}$. So $$\angle BFE = \angle BME = 90^{\circ}$$

enter image description here

Alternatively Say $AB = 2a$ and $BC= b$, then $$AF = {1\over 2}AG = {1\over 2} {4a^2\over \sqrt{4a^2+b^2}} = {2a^2\over \sqrt{4a^2+b^2}} $$ Then we have $$AF\cdot AC = {2a^2\over \sqrt{4a^2+b^2}}\cdot \sqrt{4a^2+b^2} = 2a^2 = AM\cdot AB$$ and so $B,C,F,M$ are concylic.

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