Be $E,F,K , L,$ points in the sides $AB,BC,CD,DA$ of a square $ABCD$, respectively. Show that if $EK$ $\perp$ $FL$ , then $EK=FL$

Question

Be $E,F,K , L,$ points in the sides $AB,BC,CD,DA$ of a square $ABCD$, respectively. Show that if $EK$ $\perp$ $FL$ , then $EK=FL$.

I need help proving something like this:

Any hints?

Edited: I wrote the principal statement wrong, now it's correct.

I saw a question in this forum about a similar problem, the problem was like this:

And i thought that i can modify the problem like the first image, and after drawing a lot of squares in geogebra, i think it's true, but i don't know how to prove it.

Answer 1

As you mention, your problem is equivalent to the later problem. See the picture below, where $AG\parallel FL$ and $BH\parallel EK$.

Answer 2

Let $ABCD$ be our square, $L\in AD$, $F\in BC$, $K\in DC$ and $E\in AB$ such that $LF\perp KE.$

Now, let $R$ be a rotation by $90^{\circ}$ of the plane around the center of the square.

Let $R(L)=L'$ and $R(F)=F'.$

Thus, $L'F'||KE$ and since $KF'||EL',$ we obtain $$KE=F'L'=FL.$$

Answer 3

In a unit square draw two perp lines with an arbitrary slope $\alpha$ from its corners.

Both these inside lines $(DE,LF)$ have equal length $L=\sec \alpha$

Length $L$ remains constant only by arbitrary parallel translation of line segments $DE,LF$ along $(x,y)$ arrow directions shown within the square and restricted to vertical/horizontal sides of square.