证明：在直角三角形ACM中,由勾股定理,得,AM^2=AC^2+CM^2在直角三角形BMN中,由勾股定理,得,MN^2=BM^2-BN^2在直角三角形AMN中,由勾股定理,得,AN^2=AM^2-MN^2=(AC^2+CM^2)-(BM^2-BN^2)=AC^2+CM^2-BM^2+BN^2因为AM是BC边...