过 D 做 ED ⊥ DC ,ED 交 BC 于 E 因为 DC⊥AC,所以 DE ‖AC D 为AB的中点,所以 DE 是中位线 DE = AC/2 AC = 2 DE tan角BCD=1/3 ,即 DE/DC = 1/3 DC/DE = 3 tanA = DC/AC = DC/(2DE) = (DC/DE) /2 = 3/2
过 D 做 ED ⊥ DC , ED 交 BC 于 E 因为 DC⊥AC, 所以 DE ‖AC D 为AB的中点, 所以 DE 是中位线 DE = AC/2 AC = 2 DE tan角BCD=1/3 , 即 DE/DC = 1/3 DC/DE = 3 tanA = DC/AC = DC/(2DE) = (D...