Question

已知在三角形ABC中,角B等于2角C,AD垂直BC于D,M为BC的中点.求证:DM等于二分之一AB

Answer 1

取AB中点N,连接DN、MN.

因为,MN是△ABC的中位线,

所以,MN‖AC,

可得：∠DMN = ∠C .

因为,DN是Rt△ABD斜边上的中线,

所以,DN = BN = (1/2)AB ,

可得：∠BDN = ∠B .

因为,∠DNM = ∠BDN-∠DMN = ∠B-∠C = ∠C = ∠DMN ,

所以,DM = DN = (1/2)AB .