∴BE=AB/2

∫CD = AB/2

∴BE=CD

Yes //CD

∴ Quadrilateral DEBC is a parallelogram

∴ Germany//BC

∴∠FBM=∠EDM

∠FMB =∠ empirical mode decomposition

∴△FBM∽△EDM

∴BM/DM=BF/DE

DE = BC

F is the midpoint of BC.

∴BF=BC/2

∴BF/DE= 1/2

∴DM=2BM

∫DB = BM+DM = 3BM

DB=9

∴BM=3

2、

∵ trapezoid ABCD is an isosceles trapezoid.

∴∠B=∠C

∫∠B = 60

∴∠C=60

∠∠2 =∠B

∴∠2=60

∵∠ 1+∠B+∠4= 180

∠ 1+∠2+∠3= 180

∴∠3=∠4

∠B=∠C

∴△ABP∽△PCE

∴AB/PC=BP/CE

AB = 4cm, BC = 7cm, BP = 5cm.

∴PC=2

∴4/2=5/CE

∴CE=2.5cm

Well, please accept it,