Answer: 56 or 70.

Solution: Let the bisector of the angle at vertex $A$ of parallelogram $A B C D$ intersect side $B C$ at point $M$, such that $B M=7$ and $C M=14$. Then $\angle B M A=\angle M A D=\angle A M B$, so triangle $A B M$ is isosceles. Therefore, $A B=B M=7, C D=A B=7, A D=B C=7+14$. The perimeter is 56.

Similarly for the case $B M=14$. The perimeter is 70.

Criterion: one case is lost - the solution is evaluated out of 3 points. Answer, answer with verification - evaluated out of 3 points.
8.3. Find the perimeter of a parallelogram if the bisector of one of its angles divides a side of the parallelogram into segments of 7 and 14.