Let the medians of the triangle $ABC$ meet at $G$. Let $D$ and $E$ be different points on the line $BC$ such that $DC=CE=AB$, and let $P$ and $Q$ be points on the segments $BD$ and $BE$, respectively, such that $2BP=PD$ and $2BQ=QE$. Determine $\angle PGQ$.
1. **Identify the given points and segments:**
   - Let the medians of triangle $ABC$ meet at $G$.
   - Points $D$ and $E$ are on line $BC$ such that $DC = CE = AB$.
   - Points $P$ and $Q$ are on segments $BD$ and $BE$ respectively, such that $2BP = PD$ and $2BQ = QE$.

2. **Determine the position of $P$ and $Q$:**
   - Since $2BP = PD$, point $P$ divides $BD$ in the ratio $2:1$.
   - Similarly, since $2BQ = QE$, point $Q$ divides $BE$ in the ratio $2:1$.

3. **Find the midpoint $M$ of $PQ$:**
   - Let $M$ be the midpoint of $PQ$.
   - Since $P$ and $Q$ divide $BD$ and $BE$ in the ratio $2:1$, $M$ is also a point that divides $BC$ in a specific ratio.

4. **Analyze the relationship between $M$ and $G$:**
   - Since $M$ is the midpoint of $PQ$ and $G$ is the centroid of triangle $ABC$, $G$ divides each median in the ratio $2:1$.
   - Therefore, $MG \parallel AB$ because $G$ is the centroid and $M$ is a point on the median.

5. **Construct point $X$ on $AB$:**
   - Let $X$ be a point on $AB$ such that $2BX = XA$.
   - This implies $BX = PM = MQ$.

6. **Establish the parallelogram $BXGM$:**
   - Since $MG \parallel AB$ and $BX = PM = MQ$, quadrilateral $BXGM$ is a parallelogram.
   - In a parallelogram, opposite sides are equal and parallel, so $MG = BX$.

7. **Determine $\angle PGQ$:**
   - Since $BXGM$ is a parallelogram, $MG = BX = PM = MQ$.
   - Therefore, $\angle PGQ = 90^\circ$ because $P$, $G$, and $Q$ form a right angle at $G$ due to the properties of the parallelogram and the centroid.

The final answer is $\boxed{90^\circ}$