Triangle $\vartriangle ABC$ has circumcenter $O$ and orthocenter $H$. Let $D$ be the foot of the altitude from $A$ to  $BC$, and suppose $AD = 12$. If $BD = \frac14 BC$ and  $OH  \parallel  BC$, compute $AB^2$.
.
1. **Identify the given information and construct necessary points:**
   - Triangle $\triangle ABC$ has circumcenter $O$ and orthocenter $H$.
   - $D$ is the foot of the altitude from $A$ to $BC$.
   - $AD = 12$.
   - $BD = \frac{1}{4} BC$.
   - $OH \parallel BC$.

2. **Construct the antipode of $H$ across $D$ and call it $H'$:**
   - Since $H$ is the orthocenter, $H'$ is the reflection of $H$ across $D$.
   - By properties of reflection, $AH = HH'$ and $HD = DH'$.

3. **Determine the length $HD$:**
   - Let $HD = m$. Since $HD = DH'$, we have $AD = 3m$.
   - Given $AD = 12$, we solve for $m$: 
     \[
     3m = 12 \implies m = 4
     \]
   - Therefore, $HD = 4$.

4. **Use the power of point theorem at $D$ with respect to the circumcircle of $\triangle ABC$:**
   - The power of point theorem states that for a point $D$ on the circle, $BD \cdot CD = AD \cdot DD'$.
   - Given $BD = \frac{1}{4} BC$, let $BC = 4x$. Then $BD = x$ and $CD = 3x$.
   - Using the power of point theorem:
     \[
     BD \cdot CD = AD \cdot DD' \implies x \cdot 3x = 12 \cdot 4
     \]
     \[
     3x^2 = 48 \implies x^2 = 16 \implies x = 4
     \]
   - Therefore, $BD = 4$ and $BC = 16$.

5. **Calculate $AB^2$ using the Pythagorean theorem:**
   - In right triangle $\triangle ABD$, we have:
     \[
     AB^2 = AD^2 + BD^2
     \]
   - Substituting the known values:
     \[
     AB^2 = 12^2 + 4^2 = 144 + 16 = 160
     \]

The final answer is $\boxed{160}$