The key to solving this problem lies in the formula for the area of a triangle:

$$
\(\text { Area }=\frac{1}{2} \times \text { Base × Height }\)
$$

1. Define the Square's Dimensions

Let the side length of the square $\(A B C D\)$ be $\(s\)$.
- The Area of the Square $\(=s \times s=s^2\)$.
- The side $A B$ is parallel to the side $C D$.
- The side $A D$ is parallel to the side $B C$.
- The length of $\(A B=s\)$.

2. Calculate the Area of the Shaded Region (Quantity A)

The shaded region is Triangle $A B E$.
- Base: We can use side $A B$ as the base. Thus, Base $=A B=s$.
- Height: The height of the triangle must be the perpendicular distance from the vertex $E$ to the line containing the base $A B$. Since $A B C D$ is a square and $E$ is on $C D$, this perpendicular distance is equal to the side length of the square, $A D$ or $B C$, which is $s$.

$$
\(\begin{gathered}
\text { Area of Shaded Region }=\text { Area of } \triangle A B E=\frac{1}{2} \times \operatorname{Base}(A B) \times \text { Height } \\
\text { Area of Shaded Region }=\frac{1}{2} \times s \times s=\frac{1}{2} s^2
\end{gathered}\)
$$

3. Calculate the Area of the Unshaded Region (Quantity B)

The unshaded region is everything in the square that is not shaded.

Area of Unshaded Region $=$ Area of Square $A B C D-$ Area of Shaded Region

Area of Unshaded Region $\(=s^2-\frac{1}{2} s^2=\frac{1}{2} s^2\)$

(Alternatively, the unshaded region is composed of two triangles: $\(\triangle B C E\)$ and $\(\triangle A D E\)$.

$\(\operatorname{Area}(\triangle B C E)=\frac{1}{2} \times B C \times C E\)$ and $\(\operatorname{Area}(\triangle A D E)=\frac{1}{2} \times A D \times D E\)$. Since $\(A D= B C=s\)$ and $\(C E+D E=s\)$, the total area is: $\(\frac{1}{2} s(C E+D E)=\frac{1}{2} s(s)=\frac{1}{2} s^2\)$.)

Since the area of the shaded triangle $A B E$ is exactly half the area of the square, and the unshaded area is the other half, the two quantities are equal.

Therefore, the relationship is The two quantities are equal.

let me know if still unclear