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Say we have a square $ABCD$. Put points $E$ and $F$ on sides $AB$ and $BC$ respectively, so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. What is $angle DNF$?

I'm inclined to say that it's a right angle because that's what it looks like from what I've drawn, but I have no idea how to proceed.

$angle DNF = 90^circ Longleftrightarrow angle BNF = angle CND$. It suffics to prove that $triangle BNF sim triangle CND$.
Well, it's trivial:
$angle NBF = angle BEC = angle NCD$ and $displaystyle frac{NB}{BF} = frac{NB}{BE} = sinangle NEB = cosangle NCB = frac{NC}{CB} = frac{NC}{CD}$. Q.E.D

Firstly We can write

(1) Oklid relation from $triangle CNB$

$h^2=m(k+x)$

(2) Pisagor relation from $triangle DPN$

$a^2=(x+k)^2+(x+k+m-h)^2$

(3) Pisagor relation from $triangle FRN$

$b^2=h^2+k^2$

(4) Pisagor relation from $triangle DCF$

$c^2=x^2+(x+k+m)^2$

if DNF triange is a right triangle,It must satisfy $a^2+b^2=c^2$.

$(x+k)^2+(x+k+m-h)^2+h^2+k^2=x^2+(x+k+m)^2$

$(x+k)^2+(x+k+m)^2-2h(x+k+m)+h^2+h^2+k^2=x^2+(x+k+m)^2$

$(x+k)^2-2h(x+k+m)+2h^2+k^2=x^2$

$xk+k^2-h(x+k+m)+h^2=0$

$xk+k^2-h(x+k+m)+m(k+x)=0$

$-h(x+k+m)+(k+m)(k+x)=0$

$frac{x+k}{x+k+m}=frac{h}{k+m}$

This result is equal to the rates of thales formula for similar triangles $triangle CRN sim triangle CBE$

Thus $a^2+b^2=c^2$ is correct for $triangle DNF$ .

Let us do it through coordinate geometry. Let $B$ be the origin. Lets fix the coordinates first. $$A = (0,a)\ B = (0,0) \ C = (a,0)\ D = (a,a)$$ Since $E$ and $F$ are equidistant from $B$, lets say $$F = (b,0)\ E = (0,b)$$ where $0 leq b leq a$. The equation of the line $CE$ is $dfrac{x}{a} + dfrac{y}{b} = 1$. The equation of $BN$ is $y = dfrac{a}{b}x$. This gives us the coordinate of $N$ as $left( dfrac{ab^2}{a^2 + b^2},dfrac{a^2b}{a^2 + b^2} right)$.

The slope of $FN$ is $m_1 = dfrac{a^2b/(a^2+b^2) - 0}{ab^2/(a^2+b^2)-b} = dfrac{a^2}{ab - a^2 - b^2}$.

The slope of $DN$ is $m_2 = dfrac{a^2b/(a^2+b^2) - a}{ab^2/(a^2+b^2)-a} = dfrac{ab - a^2 - b^2}{b^2 - a^2 - b^2} = - left(dfrac{ab-a^2-b^2}{a^2} right)$.

Hence, the product of the slopes is $m_1m_2 = -1$. Hence your conjecture is indeed correct. I am waiting for someone to post a nicer geometric argument.

HINT: Try to resort to Homothetic transformation and you're done immediately. Another way is to connect $N$ to the middle of the $DF$ (let's call that point $M$) and prove that you have there $NM$=$DM$=$MF$ (here you may resort to Apollonius' theorem). You also could resort to the cyclic quadrilaterals as another approaching way.