Brother of Japanese theorem for cyclic quadrilaterals 

I am looking for a proof of a like result as follows and Higher-dimensional generalizations?


Let $A, B, C, D$ be four point with lengths of $AB, BC, CD, DA$ are $a, b, c, d$ respectively. Let $F \in DA, E \in AB$ such that $AE$ $=AF$ $=\frac{da}{a+b+c+d}$ construct point $A'$ such that  $AEA'F$ is a rhombus (See Figure). Construct $B', C', D'$ cyclically, then:
1) Four points $A', B', C', D'$ form a parallelogram (like Varignon's theorem)
2) Four points $A', B', C', D'$ form a rectangle if $ABCD$ is convex cyclic quadrilateral (Like Japanese theorem for cyclic quadrilaterals.
Why I call this result is Brother of Japanese theorem for cyclic quadrilaterals?
Let $ABCD$ is convex cyclic quadrilateral, let $F \in DA, E \in AB$ such that $AE$ $=AF$ $=\frac{da}{a+d+BD}$ construct point $A'$ such that  $AEA'F$ is a rhombus then $A'$ is the incenter of $\triangle ABD$. Construct $B', C', D'$ cyclically, then $A', B', C', D'$ form a rectangle. This is Japanese theorem for cyclic quadrilaterals
3) Four points $A', B', C', D'$ are collinear with $A'C'$, $B'D'$ have the same midpoint if $ABCD$ is the concave cyclic quadrilaterals (Like Butterfly theorem)

See also:

*

*Vietnamese Extension of a Japanese Theorem
 A: 1) We have
$$\begin{gather*}
\vec E = \vec A + \frac{d}{a + b + c + d}\vec{AB}, \quad
\vec F = \vec A + \frac{a}{a + b + c + d}\vec{AD}, \\
\vec{A'} = \vec E + \vec F - \vec A = \frac{a\vec D + b\vec A + c\vec A + d\vec B}{a + b + c + d}, \\
\end{gather*}$$
and similarly
$$\begin{gather*}
\vec{B'} = \frac{a\vec C + b\vec A + c\vec B + d\vec B}{a + b + c + d}, \quad
\vec{C'} = \frac{a\vec C + b\vec D + c\vec B + d\vec C}{a + b + c + d}, \quad
\vec{D'} = \frac{a\vec D + b\vec D + c\vec A + d\vec C}{a + b + c + d}, \\
\vec{A'B'} = \frac{c\vec{AB} - a\vec{CD}}{a + b + c + d} = -\vec{C'D'}, \quad
\vec{B'C'} = \frac{d\vec{BC} - b\vec{DA}}{a + b + c + d} = -\vec{D'A'}, \\
\end{gather*}$$
so $A'B'C'D'$ is a parallelogram.
2) If $ABCD$ is convex cyclic, then $\angle ABC + \angle CDA = \angle BCD + \angle DAB = 180^\circ$, so
$$\begin{gather*}
c\vec{AB} \cdot d\vec{BC} = -abcd \cos \angle ABC = abcd \cos \angle CDA = -a\vec{CD} \cdot b\vec{DA}, \\
a\vec{CD} \cdot d\vec{BC} = -abcd \cos \angle BCD = abcd \cos \angle DAB = -c\vec{AB} \cdot b\vec{DA},
\end{gather*}$$
whence $\vec{A'B'} \cdot \vec{B'C'} = 0$ and the parallelogram is a rectangle.
3) If $ABCD$ is concave cyclic, then $\angle ABC = -\angle CDA$ and $\angle BCD = -\angle DAB$, so
$$\begin{gather*}
c\vec{AB} \times d\vec{BC} = abcd \sin \angle ABC = -abcd \sin \angle CDA = -a\vec{CD} \times b\vec{DA}, \\
a\vec{CD} \times d\vec{BC} = abcd \sin \angle BCD = -abcd \sin \angle DAB = -c\vec{AB} \times b\vec{DA},
\end{gather*}$$
whence $\vec{A'B'} \times \vec{B'C'} = 0$ and the parallelogram degenerates to a line.
