In the parallelogram $ABCD$, the angle bisector of $\angle ABC$ intersects the side $AD$ at point $P$. We know that $PD=5$, $BP=6$, $CP=6$. What is the length of side $AB$?
Given $AB < BC$, the angle bisector intersects the side $AD$ at an internal point. From the conditions of the problem, we have $\angle ABP = \angle PBC = \angle PCB = \alpha$.

In the parallelogram $ABCD$, $\angle BAD = 180^\circ - 2\alpha$. Using this, we find that $\angle APB = \alpha$. From the similar triangles $\triangle APB$ and $\triangle PBC$,

\[
\frac{AB}{BP} = \frac{BP}{BC}
\]

Denoting the side $AB$ by $x$, we arrive at the quadratic equation $x^2 + 5x - 36 = 0$. Taking the positive value, we find that the side $AB$ is 4 units.

Few considered the other case, where the angle bisector intersects the extension of the side. Using a similar reasoning, we find that $AB = 9$.

The problem can also be solved in other ways, for example, by using trigonometric functions, but the solution using similarity is the simplest.