C-14-Q030mediumsingle_mcqIn △ABC\triangle ABC△ABC, DE∥BCDE \parallel BCDE∥BC, AB=9AB = 9AB=9, AD=3AD = 3AD=3, AC=12AC = 12AC=12. Then AE=?AE = ?AE=?a333b444c666d99ব্যাখ্যাSince DE∥BCDE \parallel BCDE∥BC, the Basic Proportionality Theorem gives ADAB=AEAC\dfrac{AD}{AB} = \dfrac{AE}{AC}ABAD. Substituting, , so .