Geometry problem about square

Question

We need to solve this problem without using trigonometry.

KLMN square is inside ABCD square. Prove that midpoints of the segments AK, BL, CM, and DN are vertices of a square. KLMN can be situated anywhere inside ABCD. I think that we are going to use congruent triangles.

I tried to draw so that the intersection point of diagonals of ABCD and KLMN coincide, I also tried the opposite to consider the general case, because KLMN can take an arbitrary position.

So, I would like to see approaches, how to prove.

Answer 1

COMMENT.-Let $S$ and $s$ be the sides of both squares.

$(1)$ Inside square $ABCD$ take a point $N$.

$(2)$ Draw the circle $\Gamma$ centered at $N$ having radius $s$.

$(3)$ Take a point $K$ in $\Gamma$.

$(4)$ The line $NM$ perpendicular to side $NK$, cuts $\Gamma$ in $M$.

$(5)$ The two lines, parallel to $NM$ and $NK$ passing by $K$ and $M$ respectively interset at point $L$ which clearly is the fourth vertex of the small square $KLMN$.

$(6)$ What remains to finish is taking the midpoints of segments $AK,BL,CM$ and $DN$ and verify that them are the vertices of an square which is easy and straightforward.

Answer 2

Hints towards a solution. If you're stuck, explain what you've tried.

• Show that the angles between the corresponding sides of the squares $KLMN$ and $ABCD$ are the same. Let the midpoints of the segments be $W,X,Y, Z$.
• Show that $WX$ is the median of a triangle with edges $AB$ and $KL$, and the angle between them is the above angle.
  • As an example, see this problem for the situation when the angle between $AB$ and $KL$ is a right angle.
  • Hence, by congruency (no need for trigo), $WXYZ$ is a rhombus.
• Show that $WX$ and $XY$ are perpendicular.
  • Think about what it means to shift $KL$, $KN$ to start at $X$, and do some angle chasing (no need for trigo).
  • Hence, $WXYZ$ is a square.

Answer 3

According to the way I read the problem, $KLMN$ can be rotated any amount when it is placed inside $ABCD$. For example, $K$ could be near $C$, $L$ near $D$, and so forth so that the square formed by the midpoints of $AK$ and the other corresponding segments is inside $KLMN$ and small.

But the orientation of $KLMN$ must be the same as $ABCD$, for example if $ABCD$ is a list of vertices in counterclockwise order then $KLMN$ must also be in counterclockwise order. It is easy to come up with counterexamples if the squares are allowed to have opposite orientation.

Given $KLMN$ with the same orientation as $ABCD$, and quadrilateral $WXYZ$ with vertices at the midpoints of $AK$, $BL$, $CM$, and $DN$, you might show that translating $KLMN$ without moving $ABCD$ leads to the quadrilateral $WXYZ$ being translated half as much without any change in shape or size.

You could translate $KLMN$ to $K'L'M'N'$ so that $K'$ coincides with $A$. In most cases this would cause part of $KLMN$ (or maybe all of it) to fall outside $ABCD$. But now you can easily use congruent triangles to prove that the midpoints of $AK'$, $BL'$, and $DN'$ form two equal segments at right angles.

Translate back to the original $KLMN$ and now you know that $ZW$ and $WX$ are equal segments at right angles.

But there is nothing special about $K$ and $A$ that you can't do with $L$ and $B$, $M$ and $C$, or $N$ and $D$. The same method shows that $WX$ and $XY$ are equal segments at right angles, so are $XY$ and $YZ$, and so are $YZ$ and $ZW$.

Therefore all sides of $WXYZ$ are of equal length and all angles are right angles, so it is a square.