The bisector of an angle of a parallelogram cuts off an isosceles triangle from it.

## Solution

Let the bisector of angle $A$ intersect side $B C$ at point $P$, and line $C D$ at point $M$. Denote $A B = C D = a$. Then $B C = A D = 2a$. Since

$\angle B P A = \angle D A P = \angle B A P$, triangle $A B P$ is isosceles. Therefore, $B P = A P = a$, and $P C = B C - B P = 2a - a = a$.

Triangles $P M C$ and $P A B$ are congruent by side and the adjacent angles, so $M C = A B = a$. Similarly, $D N = a$. Therefore, $M N = M C + C D + D N = a + a + a = 3a = 12$, from which $a = 4$.

## Answer

$4, 8, 4, 8$.
[ Properties and characteristics of a parallelogram ] [ Congruent triangles. Criteria for congruence ]

Side $BC$ of parallelogram $ABCD$ is twice as long as side $AB$. The bisectors of angles $A$ and $B$ intersect line $CD$ at points $M$ and $N$, respectively, and $MN=12$.

Find the sides of the parallelogram.