Consider a $ABCD$ center square $ O $. Let $E,F,G$ and $H$ points inside the sides $AB,BC,CD$ and $DA$, respectively, such that $AE = BF =CG = DH$. It is known that $OA$ intersects $HE$ at point $X,OB$ intersects $EF$ at point $Y, OC$ intersects $FG$ at point $Z$ and $OD$ intersects $GH$ at point $W$. Since Area $EFGH = 1$, calculate $(\text{Area}$ $ABCD) \times (\text{Area}$ $XYZW)$
Attemp: WLOG, let $E, F, G, H$ be midpoints. From symmetry, $X, Y, W, Z$, will be nice midpoints too. Since each side of $EFGH=1$, we have that $BE=\frac{1}{\sqrt2}$ (and similar sides). This means that each side of $ABCD=\frac{2}{\sqrt2}$. The area of $ABCD=2$. Since $X, Y, W, Z$ are the midpoints of $EFGH$, $EX=1/2$. This means that $XY=\frac{\sqrt2}{2}$ (and other similar sides). The area of $XYWZ=\frac{1}{2}$. The answer is $1$.
Correct? Can you generalize?
Consider the points close to corner $B$ and side $AB$. (Other cyclic cases are similar) From all the rotational symmetries,
$$\begin{align*} AB &\parallel XY\\ \angle YEB &= \angle XYE &&\text{(alternate angles)}\\ \angle AEX &= \angle EXY &&\text{(alternate angles)}\\ \angle BFY &= \angle AEX &&\text{(symmetry)}\\ \triangle BFE &\sim \triangle EXY&&\text{(AAA)} \end{align*}$$
Then consider the ratio of sides of squares $ABCD$, $EFGH$ and $XYZW$,
$$\begin{align*} \frac{AB}{EF} &= \frac{AE+EB}{EF}\\ &= \frac{BF + EB}{EF} &&\text{(}AE=BF\text{ given)}\\ &= \frac{BF}{EF} + \frac{EB}{EF}\\ &= \frac{EX}{YX} + \frac{YE}{YX} &&\text{(corresponding sides)}\\ &= \frac{FY}{YX} + \frac{YE}{YX} &&\text{(}EX=FY\text{ symmetry)}\\ &= \frac {EF}{XY}\\ \end{align*}$$
So the side lengths of the three squares are in geometric sequence, so their areas also are in geometric sequence.
$$ \begin{align*} \frac {(AB)^2}{(EF)^2} &= \frac {(EF)^2}{(XY)^2}\\ (AB)^2(XY)^2 &= (EF)^4\\ &= 1 \end{align*}$$