Why is this problem still unsolved? Let $ X = AC\cap BD$ and $ Y = A_1C_1\cap B_1D_1$ Then it is well known that $ X = Y$ We also known $ EFGH$ is always parallelogram. Now $ EFGH$ is rectangel $ \Leftrightarrow$ $ A_1C_1\bot B_1D_1$. Let $ T = AB\cap CD$ Now angle chase ...

Since $A_1C_1||EF$ and $B_1D_1||EH$, we have: $\angle BAD=\pi-2\angle AA_1D_1=\pi-2\angle A_1B_1D_1=\pi-2\angle A_1EH$ $\angle BCD=\pi-2\angle CB_1C_1=\pi-2\angle B_1A_1C_1=\pi-2\angle B_1EF$ Suppose $EFGH$ is a rectangle. So we have $\angle FEH=\frac{\pi}{2}$, therefore: $\angle A_1EH+\angle B_1EF=\frac{\pi}{2}\Leftrightarrow-2\angle A_1EH-2\angle B_1EF=-\pi\Leftrightarrow$ $\Leftrightarrow(\pi-2\angle A_1EH)+(\pi-2\angle B_1EF)=\pi\Leftrightarrow\angle BAD+\angle BCD=\pi$ Therefore $ABCD$ is cyclic, QED. Suppose $ABCD$ is cyclic. So we have $\angle BAD+\angle BCD=\pi$, therefore: $\angle BAD+\angle BCD=\pi\Leftrightarrow(\pi-2\angle A_1EH)+(\pi-2\angle B_1EF)=\pi\Leftrightarrow$ $\Leftrightarrow-2\angle A_1EH-2\angle B_1EF=-\pi\Leftrightarrow\angle A_1EH+\angle B_1EF=\frac{\pi}{2}$ Therefore $\angle FEH=\frac{\pi}{2}$ and analogously $\angle EFG=\angle FGH=\angle EHG=\frac{\pi}{2}$, therefore $EFGH$ is a rectangle, QED.

Invert about the inscribed circle. Now $A$ inverts to $E$, $B$ to $F$, $C$ to $G$ and $D$ to $H$. $A,B,C,D$ are cyclic if and only if $E,F,G,H$ are cyclic. Also note that $E,F,G,H$ is a parallelogram by Varigon's Theorem. The rest is trivial.

$ABCD$ concyclic suffices that, $\angle A +\angle C=\pi , $ suffices $A_1C_1 \perp B_1D_1$, suffices $EFGH$ is a rectangle.

genxium wrote: $ABCD$ concyclic suffices that, $\angle A +\angle C=\pi , $ suffices $A_1C_1 \perp B_1D_1$, suffices $EFGH$ is a rectangle. why A1C1 perpendicular to B1D1

I think it's because $\angle C_1D_1B_1 +\angle D_1C_1A_1 = \frac{\pi}{2},$ derived by $\angle A +\angle C=\pi$. BTW, the derivation can be easily reversed so that I didn't state the inverse version.