wrong sol can we use inversion perhaps (inspired by egmo)

how will u find incentres when circle is not unit??

If $I_A, I_B, I_C, I_D$ are their respective incenters then the midpoint of $I_AI_C$ lies on the line halfway between the bisectors of $\angle{A}, \angle{C}$, and the midpoint of $I_BI_D$ lies on the line halfway between the bisectors of $\angle{B}, \angle{D}$, so the centers of $I_AI_BI_CI_D, ABCD$ are the same, so the incircles centered at $I_A, I_C$ are congruent, as are the ones centered at $I_B, I_D$. Now fixing $W, Z$, (which also fixes $I_C$), if $XY \parallel WZ$ then the incircles centered at $I_B, I_D$ are congruent, and if not then one is larger than the other, so $WXYZ$ is a parallelogram.

Let $I_A,I_B,I_C$ and $I_D$ be the incenters of triangles $AWZ, BWX,CXY$ and $DZY,$ respectively. Moreover, let $O$ be the center of parallelogram $ABC$. Note that quadrilaterals $AI_AI_DD$ and $BI_BI_CC$ are equal, since they have their sides pairwise parallel ($AI_A \parallel CI_C$ and $BI_B \parallel DI_D$ due to trivial angle-chasing, $AD \parallel BC$ since $ABCD$ is a parallelogram, and $I_AI_D \parallel I_BI_C$ since $I_AI_BI_CI_D$ is a paralelogram) and they also have two equal sides, $AB$ and $CD$. Thus, we obtain that $AI_A=CI_C$ and $BI_B=DI_D$. Now, let $Z'$ and $W'$ be the reflections of $Z,W$ across $O$. Note that $I_A$ maps to $I_C$ under this reflection, and so triangles $CXY$ and $CZ'W'$ have a common incenter. Therefore, $\angle Z'I_CW'=\angle XI_CY$, and so $(Z'C-XC)(W'C-YC) \leq 0$ must hold. Indeed, if the opposite inequality were true, then either $Z'C>XY$ and $W'C>YC$, or $Z'C<XY$ and $W'C<YC$. The first case leads to $\angle CI_CY<\angle Z'I_CW'$, while the second case leads to $\angle CI_CY>\angle Z'I_CW'$. Now, assume that $Z'C>XC$ was true. Then, $W'C \leq YC$, and so we obtain $AZ>XC$ and $AW \leq YC$. Thus, $DZ<BX$, and in a similar manner (reflecting $Z,Y$ across $O$) this implies that $DY \geq BW$, and so $AW \geq YC$, which contradicts $AW \leq YC$, unless $AW=YC$. However, in this case points $W'$ and $Y$ coincide, and since trianges $CXY$ and $CW'Z'$ have a common incenter, points $Z'$ and $X$ must coincide too, which implies that $AZ=CX$, a contradiction. Similarly, we reach a contradiction if $Z'C<XC$ was true. Therefore, $XC=Z'C=AZ,$ and similarly we obtain that $AW=YC$. Thus, $AWCY$ is a parallelogram and so $WY$ bisects $AC$. Similarly $ZX$ bisects $BD,$ and so $WY,ZX,AC$ and $BD$ concur at the center $O$, which obviously finishes the problem.

hahahahahahahahaha what Let $I_A,\ldots,I_D$ be the incenters and $r_1,\ldots,r_4$ be the radii of $I_A,\ldots,I_D$ respectively. Since $\overline{I_AI_B}$ and $\overline{I_CI_D}$ have the same "slope" with respect to $AB \parallel CD$ and are the same length, it follows that $$r_1-r_2=d(I_A,\overline{AB})-d(I_B,\overline{AB})=d(I_C,\overline{BC})-d(I_D,\overline{BC})=r_3-r_4,$$so $r_1+r_4=r_2+r_3$. Likewise, considering $\overline{I_BI_C}$ and $\overline{I_DI_A}$, $r_1+r_2=r_3+r_4$. Furthermore, these four sums must actually equal the same quantity: $\tfrac{r_1+r_2+r_3+r_4}{2}$, so we readily obtain $r_1=r_3$ and $r_2=r_4$. WLOG let $\angle B$ be non-acute. Fix $r_1$ and the positions of $W$ and $Z$, and let $X$ vary along $\overline{BC}$, with $Y$ following such that $r_3$ is fixed. I will show that there is a unique position for $X$ (and therefore $Y$) such that $r_2=r_4$. Clearly the position of $Y$ is uniquely determined, and as $X$ gets closer to $B$, $Y$ gets closer to $C$. Furthermore, as this happens (i.e. $BX$ decreases), by the sine area formula $[BXW]$ decreases and by the law of cosines $XW$ decreases, since $\cos \angle B\leq 0$, so the semiperimeter of $\triangle BXW$ decreases too, so the value of $r_2$ decreases. Analogously, as $CY$ increases, the value of $r_4$ increases. Therefore there can be at most one choice of $\overline{XY}$ such that $r_2=r_4$. However, by symmetry it is clear that making $\overline{XY}$ the image of $\overline{WZ}$ under a $180^\circ$ rotation about the center of $ABCD$ works, so we must have $AZ=CX$ and $AW=CY$, from which it is clear (by vectors, to be explicit) that the midpoints of $\overline{WY}$ and $\overline{XZ}$ are the center of $ABCD$, hence $WXYZ$ is a parallelogram. $\blacksquare$ Remark: this is an algebra and combinatorics problem

Denote center of $\triangle{AWZ},\triangle{BXW},\triangle{CYX},\triangle{DZY}$ as $I_{1},I_{2},I_{3},I_{4}$ respectively with radius $r_{1},r_{2},r_{3},r_{4}$ we Claim: $r_{1}=r_{3}$ and $r_{2}=r_{4}$ Pf:- denote $AI_{1} \cap BI_{2}=P$ and $CI_{3}\cap DI_{4} =Q$ we have $\triangle{CDQ} \cong \triangle{ABP}$ and also since $I_{1}I_{2}I_{3}I_{4}$ is a $\parallel$ gm we have $I_{1}I_{2}=I_{3}I_{4}$ and also we have $AEP \parallel CQ$ which gives $\angle{PI_{1}I_{2}}=\angle{I_{4}I_{3}Q}$ and $\angle{PI_{2}I_{1}}=\angle{QI_{4}I_{3}}$ which gives $\triangle{I_{4}I_{3}Q} \cong \triangle{I_{2}I_{1}P} \implies PI_{1}=QI_{3}$ and also we have $PA=QC$ hence we have $AI_{1}=CI_{3}$ or $\frac{r_{1}}{\sin\left(\frac{\theta}{2}\right)}=\frac{r_{3}}{\sin\left(\frac{\theta}{2}\right)} \implies r_{1}=r_{3}$ where $\angle{DAB}=\angle{BCD}=\theta$ in similar fashion we have $r_{2}=r_{4}$ so claim is done. $\square$ now from claim we have that incircle of $\triangle{AZW}$ and $\triangle{CYX}$ is symmetric about $BD$ hence if $W.L.O.G$ we consider $AZ>CX$ we have $AW<CY$ and which implies $BW>DY$ in order to have $AB=CD$ but we also have incircle of $\triangle{DZY}$ and $\triangle{BXW}$ to be symmetric about $AC$,hence we have $DZ>BX$ and that gives $AZ<CX$ in order to have $AD=BC$ but this is a contradiction .$\Rightarrow\!\Leftarrow$ hence we have $AZ=CX$ and $BW=DY$ which gives $\triangle{WAZ} \cong \triangle{XCY}$ so we have $ZW=CX$ and similarly $ZY=WX$ which gives $WXYZ$ is a parallelogram. $\square$

[asy][asy] size(10cm); import olympiad; import geometry; pair A = (0,0); pair B = (4,0); pair D = (1,3); pair C = B+D-A; pair W = (1.7,0); pair X = 0.3*B + 0.7*C; pair Y = B+D-W; pair Z = B+D-X; pair I_1 = incenter(A,W,Z); pair I_2 = incenter(B,W,X); pair I_3 = incenter(C,X,Y); pair I_4 = incenter(D,Y,Z); pair J_1 = I_1+B-A; pair J_4 = I_4+B-A; fill(B--J_1--I_2--cycle, paleblue); fill(C--I_3--J_4--cycle, paleblue); draw(A--B--C--D--cycle, linewidth(1)); draw(W--X--Y--Z--cycle, red+linewidth(0.8)); draw(I_1--I_2, blue+linewidth(0.8)); draw(I_3--I_4, blue+linewidth(0.8)); draw(I_1--I_4, blue+linewidth(0.8), StickIntervalMarker(1, 2, p=blue+linewidth(0.8), size=0.3cm)); draw(I_2--I_3, blue+linewidth(0.8), StickIntervalMarker(1, 2, p=blue+linewidth(0.8), size=0.3cm)); draw(incircle(A,W,Z), red); draw(incircle(B,W,X), red); draw(incircle(C,X,Y), red); draw(incircle(D,Y,Z), red); draw(B--J_1--I_2--cycle, blue+linewidth(0.8)); draw(C--I_3--J_4--cycle, blue+linewidth(0.8)); draw(J_1--J_4, blue+linewidth(0.8), StickIntervalMarker(1, 2, p=blue+linewidth(0.8), size=0.3cm)); dot("$A$", A, dir(-145)); dot("$B$", B, dir(-73)); dot("$D$", D, dir(125)); dot("$C$", C, dir(62)); dot("$W$", W, dir(-90)); dot("$X$", X, dir(-19)); dot("$Y$", Y, dir(90)); dot("$Z$", Z, dir(161)); dot("$M$", (B+D)/2, dir(90)); dot("$I_1$", I_1, dir(-145)); dot("$I_2$", I_2, dir(-116)); dot("$I_3$", I_3, dir(109)); dot("$I_4$", I_4, dir(125)); dot("$J_1$", J_1, dir(-40)); dot("$J_4$", J_4, dir(5)); [/asy][/asy] Let $M$ be the center of the parallelogram. Let $I_1, I_2, I_3, I_4$ be the incenters of $\triangle AWZ$, $\triangle BXW$, $\triangle CXY$, and $\triangle DYZ$. Let $J_1 = I_1 + B-A$ and $J_4 = I_4 + B-A$. Then, we see that $I_1I_4$, $I_2I_3$, and $J_1J_4$ are parallel and equal. Thus, $I_2J_1J_4I_3$ is a parallelogram, implying that $I_2J_1$ and $I_3J_4$ are parallel and equal. However, notice that $CI_3\parallel AI_1\parallel BJ_1$ because they are angle bisectors of two sides of the parallelogram. Similarly, $CJ_4\parallel BI_2$, so $\triangle BI_2J_1\cong\triangle CJ_4I_3$, implying that $BJ_1 = CI_3$. Thus, $AI_1=CI_3$, implying $M$ is also the center of $I_1I_2I_3I_4$. Now, let $W'$ and $Z'$ be the reflections of $W$ and $Z$ across $M$. By the argument above, $W'Z'$ is also tangent to the incircle of $\triangle CXY$. Assume that $Z'$ lies strictly inside segment $\overline{CX}$ (the other case, $W'$ lies inside segment $\overline{CY}$ follows similarly). Then, we see that $$\operatorname{inradius}(\triangle DZY) > \operatorname{inradius}(\triangle DW'Z) = \operatorname{inradius}(\triangle BWZ') > \operatorname{inradius}(\triangle BWX), $$a contradiction.

Here's a way to finish that does not require geometric continuity arguments or inequalities: Let $AB = b, BC = a$. Let $AW = w, BX = x, CY = y, DZ = z$. Let the incenters of $AWZ, BXW, CYX, DZY$ be $I_A, I_B, I_C, I_D$. Finally, let $ZW = p, WX = q, XY = r, YZ = s$. Let $P^l$ denote the projection of point P onto line l. Then: $I_A^{AB}I_B^{AB} = I_C^{CD}I_D^{CD}$ which implies that: $$ \frac{r + y - (a-x)}{2} + \frac{(b-y) + z -s}{2} = \frac{(q+ (b-w) - x}{2} + \frac{p + w - (a- z)}{2}$$This reduces to: $$x-z = \frac{p+q- r- s}{2}$$Similarly: $$w-y = \frac{p+s-q-r}{2}$$As a result, CXY has the same perimeter as AWZ. Now let $X', Y'$ be the reflections of X, Y about the center of ABCD, so that $X' \in AD, Y' \in AB$. Then, $AWZ$ and $AX'Y'$ have the same excircle (since they have the same perimeter). They also have the same incircle (this can be proved as in any of the above solutions). This means $X'Y'$ either coincides with $WZ$ or is its reflection about the angle bisector of $\angle A$. In either case $XY = X'Y' = WZ$. Similarly, $WX = YZ$. Thus, $WXYZ$ is a parallelogram.

since the midpoints of $I_aI_c$ and $I_bI_d$ (which are actually the same point) are on lines equidistant from $(AI_a;CI_c)$ and $(BI_b;DI_d)$ then the center of $I_aI_bI_cI_d$ is the same as that of $ABCD$, hence $AC$ and $I_aI_c$ have the same midpoint from where $AI_a \equiv CI_c$. (the following was wrong, is currently incomplete)

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Triangle AWZ and CYX is similar Triangle DZY and BXW is similar XY and WZ is parallel WX and ZY is parallel Triangle WZX and XYZ is similar ZYW and XYW triangles are also similar Therefore, it proves that WXYZ is a parallelogram

You can't know that those triangles are similar, I was wrong

Am I wrong?

Let $I_1,I_2,I_3,I_4$ be the incenters of $\triangle AZW,\triangle BXW,\triangle CYX,\triangle DZY$, respectively. Let $E=AC\cap BD$ be the intersection of diagonals. Note that $E$ is the mid-point $\overline{AC}$ and $\overline{BD}$. Note that reflecting about $E$, $A$ goes to $C$ and $B$ goes to $D$. Moreover, the angle-bisectors of $\angle BAD$ and $\angle DCB$ are parallel, so they must go to each other. Let $E'=I_1I_3\cap I_2I_4$ be the intersection of diagonals of $I_1I_2I_3I_4$. Then, reflection about $E'$, angle bisectors of $\angle BAD$ and $\angle DCB$ go to each other, as they are parallel and $I_1$ goes to $I_3$. Suppose $E,E'$ are distinct points. Then, $EE'$ is the mid-line of parallel lines $\overline{I_1I_2},\overline{I_3I_4}$. So, $\overline{EE'}\parallel\overline{I_1I_2}$. Similarly, $\overline{EE'}\parallel\overline{I_2I_3}$, which means $\overline{I_1I_2}\parallel\overline{I_2I_3}$, and hence $I_1,I_2,I_3,I_4$ are collinear, which is impossible. Hence, $E\equiv E'$ is forced. So, $I_1$ goes to $I_3$ and $I_2$ goes to $I_4$ under reflection about $E$. So, $AI_1=CI_3$, so the incircles of $AWZ$ and $CYX$ have same radii. So, incircle of $AWZ$ is reflection of incircle $CYX$ about $E$. Lemma. If $\triangle ABC$ and $\triangle PQR$ have the same inradii, and $\angle BAC=\angle QPR$, then: $AB=PQ\Longleftrightarrow AC=PR$. if $AB\neq PQ$ then $(AB-PQ)(AC-PR)<0$, i.e., it is not possible that $AB>PQ$ and $AC>PR$ simultaneously hold. Proof. First condition is trivial. For second condition, note that if any side $AB$ or $AC$ is increased, the opposite angle is increased, and the length $AI$ is increased, which is directly proportional to the inradius. So, inradius also increases. Hence, if both sides are increased, the inradius increases. Similarly, if both sides are decreased, then inradius also decreases. $\blacksquare$ We now claim that $AW=YC$. WLOG suppose $AW<YC$. Then, from our lemma, \[AZ>XC\Longrightarrow ZD=AD-AZ<BC-XC=BX.\]Again from our lemma $DY>WB$ and we also had $YC>AW$, adding gives \[DC=DY+YC>AW+WB>AB,\]which is absurd. A similar argument $AW=YC$, $ZD=BX$ and $DY=WB$. So, \[\triangle AWZ\cong\triangle CYX,\triangle BXW\cong\triangle DYZ\]from side-angle-side congruence. Note, $\angle AWZ=\angle CYX$ and $AW\parallel CY$, which means $ZW\parallel YX$. Similarly, $ZY\parallel WX$. Hence, $WXYZ$ is a parallelogram. $\blacksquare$

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