AD∑BC, e and f are the midpoint of diagonal DB and AC, respectively. Let DC midpoint G be connected with EG, then EG is the center line of △DBC, ∴EG∥BC is connected with FG, then FG is the center line of △CDA, ∴FG∥AD, ∴FG.

Step two:

Extending fe to ab at H, ∫Hg is the center line of abcd ∴hg=ad+bc/2= 12, ∫fg is the center line of △adc ∴fg=ad/2=3. Similarly, it can be proved that eh is ad/2=3, ∴.