In trapezoid $ABCD$ with $AD\parallel BC$, $AB=6$, $AD=9$, and $BD=12$. If $\angle ABD=\angle DCB$, find the perimeter of the trapezoid.
1. Given the trapezoid \(ABCD\) with \(AD \parallel BC\), we know that \(AB = 6\), \(AD = 9\), and \(BD = 12\). We are also given that \(\angle ABD = \angle DCB\).

2. Since \(\angle ABD = \angle DCB\) and \(\angle ADB = \angle DBC\) (because \(AD \parallel BC\)), triangles \(\triangle ABD\) and \(\triangle DCB\) are similar by the AA (Angle-Angle) similarity criterion.

3. Using the similarity of \(\triangle ABD\) and \(\triangle DCB\), we can set up the following ratio of corresponding sides:
   \[
   \frac{DB}{CB} = \frac{AB}{DC} = \frac{AD}{DB}
   \]

4. Substituting the given lengths into the ratio:
   \[
   \frac{AD}{DB} = \frac{9}{12} = \frac{3}{4}
   \]

5. Let \(DC = x\) and \(CB = y\). From the similarity ratio, we have:
   \[
   \frac{AB}{DC} = \frac{6}{x} = \frac{3}{4}
   \]
   Solving for \(x\):
   \[
   6 \cdot 4 = 3 \cdot x \implies 24 = 3x \implies x = 8
   \]
   Thus, \(DC = 8\).

6. Similarly, using the ratio for \(DB\) and \(CB\):
   \[
   \frac{DB}{CB} = \frac{12}{y} = \frac{3}{4}
   \]
   Solving for \(y\):
   \[
   12 \cdot 4 = 3 \cdot y \implies 48 = 3y \implies y = 16
   \]
   Thus, \(CB = 16\).

7. Now, we can find the perimeter of the trapezoid \(ABCD\) by summing all the side lengths:
   \[
   AB + AD + DC + CB = 6 + 9 + 8 + 16 = 39
   \]

The final answer is \(\boxed{39}\).