The bisectors of angles A and B of the parallelogram ABCD intersect at point K. Find the area of the parallelogram if BC = 2

The bisectors of angles A and B of the parallelogram ABCD intersect at point K. Find the area of the parallelogram if BC = 2, and the distance from point K to side AB is 1

Denote the intersection points of the bisectors with the sides as shown in the figure.
∠FAK = ∠BEK (as these are cross-angles).
It turns out that ∠BAK = ∠BEK, therefore the triangle ABE is isosceles (by the property of an isosceles triangle).
Then AB = BE.
Triangles ABK and EBK are equal by the first sign of equality of triangles.
Therefore, the heights of these triangles are also equal.
Similarly, the triangles ABK and AFK are equal.
It turns out that the height of the parallelogram is 2h.
The parallelogram area is SABCD = 2h * BC = 2 * 1 * 2 = 4
Answer: SABCD = 4