Please help me

We have triangle $\triangle ABC$ where $AB = AC$ and $AD$ is an altitude.
Meanwhile, $E$ is a point on $AC$ such that $AB \parallel DE$.
If $BC = 12$ and the area of  $\triangle ABC$ is $180$, what is the area of $ABDE$?

$\text{Let $AB = AC $ } \\ \text{Let $DB = CD =\dfrac{BC}{2} = 6$ }$

$\begin{array}{|rcll|} \hline \text{area of } \triangle ABC =180 &=& \dfrac{BC*AD}{2} \\ 180 &=& \dfrac{12*AD}{2} \\ 180 &=& 6AD \\ \mathbf{ AD } &=& \mathbf{30} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \dfrac{CD}{ED} &=& \dfrac{BC}{AB} \\\\ \dfrac{6}{ED} &=& \dfrac{12}{AB} \\\\ \dfrac{ED}{6} &=& \dfrac{AB}{12} \\\\ \mathbf{ED }&=& \mathbf{\dfrac{AB}{2}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \text{area of } ABDE &=& \dfrac{AB+ED}{2}\times H \quad| \quad H=\dfrac{DB*AD}{AB}=\dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+ED}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+\dfrac{AB}{2}}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \dfrac{3}{4}AB\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& 3* 3*15 \\\\ \mathbf{\text{area of } ABDE} &=& \mathbf{135} \\ \hline \end{array}$