在三角形ABC中,AB＞AC,AD垂直与BC于D,Q为AD上任意一点,求证：QB－QC＞AB－AC

在Rt三角形BDA中AB*AB=AD*AD+BD*BD; (1)
Rt BQD QB*QB=QD*QD+BD*BD; (2)
Rt ACD AC*AC=AD*AD+DC*DC; (3)
Rt CQD CQ*CQ=QD*QD+DC*DC; (4)
(3)-(1):AB*AB-AC*AC=BD*BD-DC*DC
AB*AB-BD*BD=AC*AC-DC*DC;(5)
(4)-(2):同理得：QB*QB-BD*BD=QC*QC-CD*CD;(6)
(5)-(6)得：AB*AB-QB*QB=AC*AC-QC*QC
AB*AB-AC*AC=QB*QB-QC*QC
(AB+AC)(AB-AC)=(QB+QC)(QB-QC)
因为AB+AC>QB+QC,
所以QB-QC>AB-AC