A is AF∨BC, and the straight line DE is at point f,

Connecting BD, it is easy to know that the quadrilateral ABEF is a rectangle, then

EF=AB=CD，∠E=90，∠BAD=∠ADF

∫≈BCD = 150 ,∴∠dce=30，

∴DE= 1/2CD= 1/2EF，

∴D is the midpoint of EF, so,

Yi Zheng delta bed delta AFD (SAS),

∴∠bde=∠adf=∠bad；

∫∠BCD = 150，BC=CD，

∴∠ EBD = 15, and ≈e = 90,

∴∠BDE=75，

∴∠BAD=75。