C-03-Q038mediumsingle_mcqIn △ABC\triangle ABC△ABC, ADADAD is the perpendicular on BCBCBC and BEBEBE is the perpendicular on ACACAC. Then —aBC⋅CD=AC⋅CEBC \cdot CD = AC \cdot CEBC⋅CD=AC⋅CEbBC⋅CD=AB⋅AEBC \cdot CD = AB \cdot AEBC⋅CD=AB⋅AEcBC⋅AC=CD⋅CEBC \cdot AC = CD \cdot CEBC⋅AC=CD⋅CEdBC+CD=AC+CEBC + CD = AC + CEBC+CD=AC+CE