In coordinate space, let $D = (0,0,1)$ and $E = (0,0,-1).$  Since $CD = EA = 2,$ $C$ lies on a circle centered at $D$ with radius 2, and $A$ lies on a circle centered at $E$ with radius 2.  Furthermore, $\angle CDE = \angle DEA = 90^\circ,$ so these circles lies in planes that are perpendicular to $\overline{DE}.$

[asy]
import three;

size(200);
currentprojection = perspective(4,3,2);

triple A, B, Bp, C, D, E;
real t;

A = (sqrt(3),1,-1);
B = (sqrt(3),-1,-1);
Bp = (sqrt(3),1,1);
C = (sqrt(3),-1,1);
D = (0,0,1);
E = (0,0,-1);

path3 circ = (2,0,-1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,-1) + (2*cos(t),2*sin(t),0));
}

draw(circ);

path3 circ = (2,0,1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,1) + (2*cos(t),2*sin(t),0));
}

draw(circ);
draw(C--D--E--A);

dot("$A$", A, S);
dot("$C$", C, W);
dot("$D$", D, NE);
dot("$E$", E, dir(0));
[/asy]

We can rotate the diagram so that $D$ and $E$ have the same $x$-coordinates.  Let $A = (x,y_1,-1)$ and $C = (x,y_2,1).$  Since $EA = CD = 2,$
\[x^2 + y_1^2 = x^2 + y_2^2 = 4.\]Then $y_1^2 = y_2^2,$ so $y_1 = \pm y_2.$

Furthermore, since $AB = BC = 2$ and $\angle ABC = 90^\circ,$ $AC = 2 \sqrt{2}.$  Hence,
\[(y_1 - y_2)^2 + 4 = 8,\]so $(y_1 - y_2)^2 = 4.$  We cannot have $y_1 = y_2,$ so $y_1 = -y_2.$  Then $4y_1^2 = 4,$ so $y_1^2 = 1.$  Without loss of generality, we can assume that $y_1 = 1,$ so $y_2 = -1.$  Also, $x^2 = 3.$  Without loss of generality, we can assume that $x = \sqrt{3},$ so $A = (\sqrt{3},1,-1)$ and $C = (\sqrt{3},-1,1).$

Finally, we are told that the plane of triangle $ABC$ is parallel to $\overline{DE}.$  Since both $A$ and $C$ have $x$-coordinates of $\sqrt{3},$ the equation of this plane is $x = \sqrt{3}.$  The only points $B$ in this plane that satisfy $AB = BC = 2$ are the vertices $B_1$ and $B_2$ of the rectangle shown below, where $B_1 = (\sqrt{3},-1,-1)$ and $B_2 = (\sqrt{3},1,1).$

[asy]
import three;

size(200);
currentprojection = perspective(4,3,2);

triple A, B, Bp, C, D, E;
real t;

A = (sqrt(3),1,-1);
B = (sqrt(3),-1,-1);
Bp = (sqrt(3),1,1);
C = (sqrt(3),-1,1);
D = (0,0,1);
E = (0,0,-1);

path3 circ = (2,0,-1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,-1) + (2*cos(t),2*sin(t),0));
}

draw(circ);
draw(surface(A--B--C--Bp--cycle),paleyellow,nolight);

path3 circ = (2,0,1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,1) + (2*cos(t),2*sin(t),0));
}

draw(circ);
draw(C--D--E--A);
draw(A--B--C--Bp--cycle);

dot("$A$", A, S);
dot("$B_1$", B, W);
dot("$B_2$", Bp, N);
dot("$C$", C, W);
dot("$D$", D, NE);
dot("$E$", E, dir(0));
[/asy]

In either case, triangle $BDE$ is a right triangle where the legs are both 2, so its area is $\frac{1}{2} \cdot 2 \cdot 2 = 2.$
좌표 공간에서 $D = (0,0,1)$, $E = (0,0,-1)$ 이므로 $CD = EA = 2,$ $C$는 반지름 2의 $D$를 중심으로 한 원에 놓이고, $A$는 반지름 2의 $E$를 중심으로 한 원에 놓입니다. 또한 $\각 CDE = \각 DEA = 90^\circ,$이므로 이 원은 $\overline{DE}.$에 수직인 평면에 놓입니다.

[asy]
세 개를 가져옵니다;

size(200);
currentprojection = perspective(4,3,2);

triple A, B, Bp, C, D, E;
실수 t;

A = (sqrt(3),1,-1);
B = (sqrt(3),-1,-1);
Bp = (sqrt(3),1,1);
C = (sqrt(3),-1,1);
D = (0,0,1);
E = (0,0,-1);

path3 circ = (2,0,-1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,-1) + (2*cos(t),2*sin(t),0));
}

draw(circ);

path3 circ = (2,0,1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,1) + (2*cos(t),2*sin(t),0));
}

draw(circ);
draw(C--D--E--A);

dot("$A$", A, S);
dot("$C$", C, W);
dot("$D$", D, NE);
dot("$E$", E, dir(0));
[/asy]

다이어그램을 회전하여 $D$와 $E$가 동일한 $x$ 좌표를 갖도록 할 수 있습니다.  A = (x,y_1,-1)$ 및 $C = (x,y_2,1)$이므로 $EA = CD = 2,$입니다.
\[x^2 + y_1^2 = x^2 + y_2^2 = 4.\]그러면 $y_1^2 = y_2^2,$이므로 $y_1 = \pm y_2.$입니다.

또한, $AB = BC = 2$이고 $\angle ABC = 90^\circ,$ $AC = 2 \sqrt{2}.$ 따라서,
\[(y_1 - y_2)^2 + 4 = 8,\]이므로 $(y_1 - y_2)^2 = 4.$ 우리는 $y_1 = y_2,$ 따라서 $y_1 = -y_2.$ 그러면 $4y_1^2 = 4,$ 따라서 $y_1^2 = 1.$ 일반성을 잃지 않고 $y_1 = 1,$ 따라서 $y_2 = -1이라고 가정할 수 있습니다.또한 $x^2 = 3.$ 일반성을 잃지 않고 $x = \sqrt{3},$ 따라서 $A = (\sqrt{3},1,-1)$ 및 $C = (\sqrt{3},-1,1)$ 이라고 가정할 수 있습니다.

마지막으로 삼각형 $ABC$의 평면은 $\overline{DE}$에 평행합니다. $A$와 $C$는 모두 $\sqrt{3}$의 $x$ 좌표를 가지므로, 이 평면의 방정식은 $x = \sqrt{3}입니다.이 평면에서 $AB = BC = 2$를 만족하는 유일한 점 $B$는 아래 표시된 직사각형의 꼭짓점 $B_1$과 $B_2$이며, 여기서 $B_1 = (\sqrt{3},-1,-1)$ 및 $B_2 = (\sqrt{3},1,1)$입니다.

[asy]
3을 가져옵니다;

size(200);
currentprojection = perspective(4,3,2);

triple A, B, Bp, C, D, E;
실수 t;

A = (sqrt(3),1,-1);
B = (sqrt(3),-1,-1);
Bp = (sqrt(3),1,1);
C = (sqrt(3),-1,1);
D = (0,0,1);
E = (0,0,-1);

path3 circ = (2,0,-1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,-1) + (2*cos(t),2*sin(t),0));
}

draw(circ);
draw(surface(A--B--C--Bp--cycle),paleyellow,nolight);

path3 circ = (2,0,1);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,0,1) + (2*cos(t),2*sin(t),0));
}

draw(circ);
draw(C--D--E--A);
draw(A--B--C--Bp--cycle);

dot("$A$", A, S);
dot("$B_1$", B, W);
dot("$B_2$", Bp, N);
dot("$C$", C, W);
dot("$D$", D, NE);
dot("$E$", E, dir(0));
[/asy]

두 경우 모두 삼각형 $BDE$는 두 다리가 모두 2인 직각 삼각형이므로 그 면적은 $\frac{1}{2}입니다. 2 \cdot 2 \cdot 2 = 2.$입니다.