Parallelogram from arbitrary quadrilateral Let $ABCD$ be an arbitrary quadrilateral, four perpendicular bisector of $AB, BC, CD, DA$ form quadrilateral $A_1B_1C_1D_1$. Let $MNPQ$ be the Varignon parallelogram of $A_1B_1C_1D_1$. Let $M', N', P', Q'$ be the reflection of $M, N, P, Q$ in $AB, DA, CD, BC$ respectively. Then show that $M'N'P'Q'$ be a parallel logram. The parallelogram is a rectangle if and only if the diagonals of the quadrilateral are perpendicular, that is, if the quadrilateral is an orthodiagonal quadrilateral.

 A: For the basic fact, we can argue very generally.
Let $\square ABCD$ and $\square MNPQ$ be arbitrary quadrilaterals, and let $\square M^\prime N^\prime P^\prime Q^\prime$ be the quadrilateral whose vertices are the reflections of the corresponding vertices of $\square MNPQ$ in the midpoints of respective segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DA}$. That is, interpreting the points as vectors, suppose we have
$$M + M^\prime = A+B \qquad 
N+N^\prime = B+C \qquad
P+P^\prime = C+D \qquad
Q+Q^\prime = D+A  \tag{$\star$}$$
Then,
$$\begin{align}
\overrightarrow{M^\prime N^\prime} + \overrightarrow{P^\prime Q^\prime} 
&= (N^\prime - M^\prime) + ( Q^\prime - P^\prime ) \\
&= (B+C-N)-(A+B-M)+(D+A-Q)-(C+D-P) \\
&= (M-N)+(P-Q) \\
&= \overrightarrow{NM} + \overrightarrow{QP}
\end{align}$$
so that

$$\begin{align}
\square MNPQ \text{ is a parallelogram} &\iff \overrightarrow{NM} + \overrightarrow{QP} = 0 \\
&\iff \overrightarrow{M^\prime N^\prime} + \overrightarrow{P^\prime Q^\prime} = 0 \\[4pt]
&\iff \square M^\prime N^\prime P^\prime Q^\prime \text{ is a parallelogram}
\end{align}$$ 


The rectangularity result requires much more specificity. It isn't enough, for instance, to require that $M$, $N$, $P$, $Q$ (and thus also $M^\prime$, $N^\prime$, $P^\prime$, $Q^\prime$) lie somewhere on the perpendicular bisectors of the sides of $\square ABCD$.
In fact, the rectangularity result is false as an "if and only if" statement. We can show this via a coordinate-based argument. Place diagonal $\overline{AC}$ on the $x$-axis, and let diagonal $\overline{BD}$ cross it at the origin, to that we have
$$A = (a,0)\qquad B=(b \cos\theta, b\sin\theta) \qquad C = (c,0) \qquad D = (d\cos\theta, d\sin\theta)$$
where $\theta = \angle AOB$. To ensure that the perpendicular bisectors have unique intersections, we take $a$, $b$, $c$, $d$, $\sin\theta$ to be non-zero, and we require $a\neq c$ and $b \neq d$.
Now, one readily calculates equations for the various perpendicular bisectors; the bisectors' intersections $A_1$, $B_1$, $C_1$, $D_1$; the intersections' midpoints $M$, $N$, $P$, $Q$; and finally the midpoints' reflections $M^\prime$, $N^\prime$, $P^\prime$, $Q^\prime$. No challenge there. From our previous work, we know that $\square M^\prime N^\prime P^\prime Q^\prime$ is a parallelogram; the figure is a rectangle if and only if it has a right angle at, say, vertex $M^\prime$, a condition encoded in this relation:
$$0 = \overrightarrow{M^\prime N^\prime}\cdot \overrightarrow{M^\prime Q^\prime} = \frac{\cos\theta\;( a - c )( b - d )\;\left(\;( a c - b d )^2 + 
   16 a b c d \sin^2\theta \;\right)}{16 a b c d \sin^2\theta}$$
Based on our assumptions, this reduces to
$$\cos\theta \; \left(\;( a c - b d )^2 + 
   16 a b c d \sin^2\theta \;\right) = 0$$
Certainly, the option $\cos\theta = 0$ corresponds to $\square ABCD$ having perpendicular diagonals. However, based on what we have described so far, there are no assurances that the second factor doesn't vanish; indeed, values $a = b = 1$, $c=-2$, $d=2$, and $\theta = \pi/6$, lead to a rectangular $\square M^\prime N^\prime P^\prime Q^\prime$ from a non-orthodiagonal $\square ABCD$.
 
The "problem" (so far as the proposed biconditional in the original question is concerned) is that the product $abcd$ can be negative. This possibility is avoidable by requiring, for instance, that $\square ABCD$ be a convex quadrilateral (so that $ac < 0$ and $bd < 0$); the reader is invited to consider alternative situations, but we'll go ahead and state things this way:

Given that $\square ABCD$ is convex,
  $$\square ABCD \text{ is orthodiagonal} \quad\iff\quad \square M^\prime N^\prime P^\prime Q^\prime \text{ is a rectangle}$$

