Question

在三角形ABC中,BD,CE分别是边AC,BA上的中线,M,N分别是BD,CE的中点,若BC=4.8CM,则MN 长为多

Answer 1

取 BC 中点 F

连接 MF NF

MF || DC,MF = DC/2 = AC/4

NF || EB,NF = EB/2 = AB/4

∠MFB = ∠C,∠NFC = ∠B

∠MFB + ∠NFC + ∠MFN = 180

∠C + ∠B + ∠A = 180

所以 ∠MFN = ∠A

因此 △MFN∽△CAB

MN/BC = NF/AB = MF/AC = 1/4

MN = BC/4 = 1.2 cm