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.