As shown in the figure, ab = AC, be ⊥ AC at point E, CF ⊥ AB at point F, be and CF intersect at point D, then ① △ Abe ≌ △ ACF; ② △ BDF ≌ △ CDE; ③ point D is on the bisector of ∠ BAC A. ①B. ②C. ①②D. ①②③

∵ be ⊥ AC in E, CF ⊥ AB in F ≁ AEB = ∠ AFC = 90 °, ∵ AB = AC, ≁ a = ∠ a, ≌ Abe ≌ ACF (① correct) ≁ AE = AF, ≁ BF = CE, ≁ be ⊥ AC in E, CF ⊥ AB in F, ≁ BDF = ∠ CDE, ≌ BDF ≌ CDE (② correct) ≌ DF = De, connect ad, ≁ AE = AF, de = DF, ad = a

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