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In a convex quadrilateral each vertex is connected by two line segments with the midpoints of the two opposite sides. In total eight line segments are drawn. Suppose that seven of them have the same length $a$. Prove that the length of the remaining segment is also $a$.

I'm considering to prove by contradiction. Assume the remaining segment is not equal to a, but I don't know how to get an impossible steps.

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Let the quadrilateral have vertexes $A$, $C$, $E$ and $G$ and let the midpoints between consecutive vertexes be $B$, $D$, $F$ and $H$, respectively.

Let the unknown length be $\overline{AD}$.

The triangles $ACF$, $EGB$ are isosceles. Hence the segment $BF$ is perpendicular to both $AC$ and $EG$ and divides those two triangles in half. Therefore the edges of the quadrilateral $AC$ and $EG$ are parallel.

Since the lengths $\overline{GH} = \overline{HA}$, $H$ is also the midpoint of the straight line that goes through $H$ and is perpendicular to the straight line that passes through the edges $AC$ and $EG$. We have a similar result for point $D$ because the lengths $\overline{CD} = \overline{DE}$. Hence we have that the segment $DH$ is parallel to the edges $AC$ and $EG$.

The triangle $CEH$ is also isosceles. Therefore, the edge $CE$ is perpendicular to the segment $DH$. Hence, the edge $CE$ is also perpendicular to the edges $AC$ and $EG$.

The triangles $DEG$ and $EGH$ are congruent by the side-angle-side condition. Which means that the lengths $\overline{DE} = \overline{GH}$. Therefore, since $DE$ is perpendicular to $EG$, so is $GH$. Hence, $HA$ is parallel to $CE$ and has the same length.

Consequently, triangles $CDH$ and $ADH$ are congruent, leading to the conclusion that $\overline{AD} = \overline{CH} = a$.

After reading all of this I wonder if there were a simpler way of getting this result...

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Some background and notation: Recall that joining the midpoints of a quadrilateral in order yields a parallelogram whose sides are parallel to corresponding diagonals of the quadrilateral. For $\square ABCD$, denote the sides of the parallelogram by $a$, $b$, $c$, $d$, with $a$ being the "far" side of the parallelogram from $A$, etc; and let $\triangle Aa$ denote the triangle with vertex $A$ and opposite side $a$, etc.

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Let $A'$ and $C'$ be the feet of perpendiculars from $A$ and $C$ onto (possibly extended) segments $a$ and $c$; further, let $A''$ be the point on $c$ such that $\overline{A'A''}\parallel\overline{AC}$. Clearly, $A''$ and $C'$ are distinct, unless (and only unless ("unlessss"?)) $\overline{AC}$ is perpendicular to $a$ (and $c$); that is, unlessss the parallelogram is a rectangle.

Now, if $\triangle Aa$ is isosceles, then $A'$ is the midpoint of $a$, and $A''$ is the midpoint of $c$; if $\triangle Cc$ is also isosceles, then $C'$ is the midpoint of $c$, hence it coincides with $A''$, making the parallelogram a rectangle. If, under these circumstances, $\triangle Bb$ is also-also isosceles, then we readily conclude that $\triangle Dd$ must be, too. (The altitudes of isosceles $\triangle Bb$ and $\triangle Dd$ coincide; if one altitude is the perpendicular bisector of $b$ (and $d$), then the other is, as well.) In other words,

If any three of $\triangle Aa$, $\triangle Bb$, $\triangle Cc$, $\triangle Dd$ are isosceles, then so is the fourth.

The result follows. $\square$

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