The bisectors of the angles A and D of the parallelogram ABCD intersect at a point lying on the side BC. Find AB if BC = 40

BC || AD (by definition of parallelogram)
∠BAE = ∠EAD (since AE is a bisector)
∠EAD = ∠BEA (as these are cross-angles)
Therefore, ∠BAE = ∠BEA
It turns out that the triangle ABE is an isosceles (by property), and AB = BE (by the definition of an isosceles triangle).
Similarly with the triangle ECD:
Since AB = CD (by the parallelogram property), it turns out that AB = BE = EC = CD.
Therefore, BE = BC / 2 = 40/2 = 20.
AB = BE = 20
Answer: AB = 20

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