⒎B
⒏C
16. solution: ∫de:EA = 2:3.
∴ Let DE = 2k and EA = 3k (k-0), then DA = DE+EA = 3k+2k = 5k.
∫EF∨AB
∴DE/DA=EF/AB
∴ab=da/de*ef=5k/(2k)*4= 10
In parallelogram ABCD, CD = AB = 10.