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*DCAB*AB-BD*BD=AC*AC-DC*DC;(5)(4)-(2):同理得：QB...