C-03-Q021mediumsingle_mcqIn △ABC\triangle ABC△ABC, ∠ACB\angle ACB∠ACB is acute and CDCDCD is the orthogonal projection of ACACAC on BCBCBC. Then —aAB2=AC2+BC2AB^2 = AC^2 + BC^2AB2=AC2+BC2bAB2=AC2+BC2+2BC⋅CDAB^2 = AC^2 + BC^2 + 2 BC \cdot CDAB2=AC2+BC2+2BC⋅CDcAB2=AC2+BC2−2BC⋅CDAB^2 = AC^2 + BC^2 - 2 BC \cdot CDAB2=AC2+BC2−2BC⋅CDdAB2=AC2−BC2AB^2 = AC^2 - BC^2AB2=AC2−BC2