Consider a square $ABCD$ with center $O$. Let $E, F, G$, and $H$ be points on the interiors of 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 $W$. Let $x$ and $y$ be the lengths of $AE$ and $AH$, respectively.

a) Given that Area $(EFGH) = 1 \text{ cm}^2$, calculate the value of $x^2 + y^2$.

b) Verify that $HX = \frac{y}{x+y}$. Then, conclude that $X, Y, Z$, and $W$ are vertices of a square.

c) Calculate

Area $(ABCD) \cdot$ Area $(XYZW)$.
Solution

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-09.jpg?height=500&width=500&top_left_y=379&top_left_x=855)

a) Let $x$ and $y$ be the lengths of $A E$ and $A H$, respectively. Given that $A H = E B$, $A E = B F$, and $\angle H A E = \angle E B F$, it follows that triangles $A E H$ and $E B F$ are congruent. Therefore,

$$
\angle H E F = 180^{\circ} - \angle H E A - \angle B E F = 180^{\circ} - \angle E F B - \angle B E F = 90^{\circ}
$$

Similarly, we can conclude that $\angle E F G = \angle F G H = \angle G H E = 90^{\circ}$. By the Pythagorean Theorem, $H E^{2} = x^{2} + y^{2}$. The same applies to the other sides of the rectangle $H E F G$, i.e., $E H = E F = F G = G H = 1$. Therefore, its area is $1 = A_{H E F G} = x^{2} + y^{2}$.

b) Since $A C$ is the angle bisector of $\angle H A E$, it follows from the Angle Bisector Theorem that

$$
\begin{aligned}
\frac{H X}{E X} & = \frac{A H}{A E} \\
\frac{H X}{H X + E X} & = \frac{A H}{A H + A E} \\
H X & = \frac{y}{x + y}
\end{aligned}
$$

Similarly, $E X = \frac{x}{x + y}$. The diagonal $B D$ is also the angle bisector of $\angle E B F$ and $\triangle E B F \equiv A H E$. Therefore, $E Y = H X = \frac{y}{x + y}$, and by analogy with the initial argument, now applied to the square $H E F G$, we can conclude that the points $X, Y, Z$, and $W$ are vertices of a square.

c) Applying the Pythagorean Theorem to triangle $E X Y$, we get

$$
\begin{aligned}
X Y^{2} & = E X^{2} + E Y^{2} \\
& = \frac{x^{2}}{(x + y)^{2}} + \frac{y^{2}}{(x + y)^{2}}
\end{aligned}
$$

Therefore, the area of the quadrilateral $X Y Z W$ is Area $(X Y Z W) = \frac{x^{2} + y^{2}}{(x + y)^{2}}$. Since Area $(E F G H) = 1$, it follows that $x^{2} + y^{2} = 1$ and that

$$
\text{Area}(A B C D) \cdot \text{Area}(X Y Z W) = (x + y)^{2} \cdot \frac{x^{2} + y^{2}}{(x + y)^{2}} = 1
$$