In triangle $ABC$, we have that $AB = AC,$ and $\overline{AD}$ is an altitude. Also, $E$ is a point on $\overline{AC}$ such that $\overline{BE} \parallel \overline{AC}.$ If $BC = 12$ and the area of triangle $ABC$ is $180,$ what is the area of $ABDE$?

Answer 1

Results from this solution are left to the reader as exercises.
[asy]

pair A,B,C,D,E;

A = origin;

C = (6,0);

B = (2,2);

D = (2,0);

E = (3,0);

draw(A--B--C--cycle);

draw(A--D);

draw(B--E);

label("$A$",A,NW);

label("$B$",B,W);

label("$C$",C,SE);

label("$D$",D,S);

label("$E$",E,S);

[/asy] $AB=AC$. Therefore, $\triangle ABC$ is isosceles with $AB = AC.$ From the problem statement, the altitude drawn from the vertex to the base divides the triangle into two congruent triangles.

So $ABDE$ is a parallelogram. Specifically, $ABDE$ is a rectangle.

Since $ABDE$ is a rectangle, its height is the same as the height. The area of $ABDE$ is then the base times the height. The height of $ABDE$ is the same as the height of $ABC$, and thus equals $AD$. The height of $ABC$ is half the length of $BC$, which is $6$.

Hence, the area of $ABDE$ is $6 \cdot AD$. Therefore, $6 \cdot AD = 180$, so $AD = 30$.

Therefore, the area of $ABDE$ is $6 \cdot AD = \boxed{180}$.