C-14-Q034mediumsingle_mcqIn △ABC\triangle ABC△ABC, if ADDB=AEEC\dfrac{AD}{DB} = \dfrac{AE}{EC}DBAD=ECAE where D∈ABD \in ABD∈AB, E∈ACE \in ACE∈AC, then —aDE∥BCDE \parallel BCDE∥BCbDE⊥BCDE \perp BCDE⊥BCc