In the parallelogram, AB=3, BC=4, the bisectors of ∠A and ∠D intersect BC at E and F respectively, then EF=?

Through E, draw EE' parallel to AB and intersect AD on E'. Triangle ABE is equal to AE'E, so AB is equal to AE', connect BE' and intersect AE at O. In the isosceles triangle ABE', AO is both the angle bisector and the height, so AO is perpendicular to BE', and the parallelogram ABEE' is right The angles are perpendicular to each other, so it is a rhombus, so BE is equal to AB is equal to 3, so E'D is equal to 1. Quadrilateral BFDE' is a parallelogram because angle BE'A is equal to angle FDA so BE' is parallel to DF, so BF is equal to E'D is equal to 1.

So EF is equal to 2