In parallelogram $A B C D$, the bisectors of the angles at side $A D$ divide side $B C$ at points $M$ and $N$ such that $B M$ : $M N=1: 5$. Find $B C$, if $A B=3$.

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Consider the case when $A M$ and $D N$ are the angle bisectors of the angles at vertices $A$ and $D$ of parallelogram $A B C D$ (left figure).

Triangle $A B M$ is isosceles because $\angle A M B = \angle M A D = \angle B A M$, so $B M = A B = 3$. Therefore, $M N =$

15. Similarly, triangle $D C N$ is also isosceles and $C N = C D = A B = 3$. Therefore, $B C = B M + M N + C N = 3 + 15 + 3 = 21$.
![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-11.jpg?height=250&width=1466&top_left_y=92&top_left_x=299)

Now, let the angle bisectors of the angles at the base $A D$ be rays $A N$ and $D M$ (right figure). Then triangles $A B N$ and $D C N$ are isosceles,

$C M = C D = A B = 3, B N = A B = 3$. Therefore, $B M = 1 / 2, B C = B M + C M = 3.5$.

## Answer

21 or $3.5$.