let $A$,$B$,$C$,$D$ be the vertices of $Q_1$. Let the midpoint of $AB$ be $M_1$, define $M_2$, $M_3$ and $M_4$ similarly. claim: The diagonals in $Q_1$ are perpendicular. Proof: By midlines, $M_1M_2//AC//M_3M_4$ and $M_1M_4//BD//M_2M_3$, so $M_1M_2M_3M_4$ is a cyclic parallelogram, which implies that it is a rectangle. Now, $AC \perp M_1M_4, M_1M_4 \perp M_1M_2$ and $M_1M_2 \perp BD$ completes the proof. Note that by perpendicularity condition/Pythagorean theorem we get $AB^2+CD^2=BC^2+AD^2$. Consider a right triangle with legs with lengths $AB$ and $CD$. The length of the hypotenuse in this triangle is the same as the length of the hypotenuse of a right triangle with legs with lengths $BC$ and $AD$. With these two triangles we can construct a cyclic quadrilateral with $2$ angles equal to $90^\circ$