C-14-Q068mediumsingle_mcqAfter △APQ≅△DEF\triangle APQ \cong \triangle DEF△APQ≅△DEF, we get ∠APQ=∠ABC\angle APQ = \angle ABC∠APQ=∠ABC, hence —aPQ⊥BCPQ \perp BCPQ⊥BCbPQ∥BCPQ \parallel BCPQ∥BCc