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 $ABDE$?
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 $6 \cdot AD = \boxed{180}$.
We are given that $AB = AC,$ so triangle $ABC$ is isosceles.
Since $\overline{AD}$ is an altitude, it bisects $\overline{BC}.$ Therefore, $BD = DC = 6.$
Since $E$ is a point on $\overline{AC}$ such that $\overline{BE} \parallel \overline{AC},$ we have that triangles $ABE$ and $ABC$ are similar.
The ratio of the areas of similar triangles is the square of the ratio of their corresponding side lengths.
Since $AB = AC,$ we have that $ABE$ and $ABC$ are congruent.
Therefore, the area of $ABDE$ is equal to half the area of triangle $ABC.$
Since the area of triangle $ABC$ is $180,$ the area of $ABDE$ is $180/2 = \boxed{90}.$