Question

AM是三角形ABC的BC边上的中线,求证AB的平方+AC的平方=2(AM的平方+BM的平方)

Answer 1

过A点作BC的垂线,交BC于D点则

AB^2=AD^2+BD^2

AC^2=AD^2+CD^2

AB^2+AC^2=2AD^2+BD^2+CD^2

AM^2=AD^2+MD^2

AD^2=AM^2-MD^2

∴AB^2+AC^2=2AD^2+BD^2+CD^2

=2(AM^2-MD^2)+BD^2+CD^2

=2AM^2+BD^2-MD^2+CD^2-MD^2

=2AM^2+(BD-MD)(BD+MD)+(CD-MD)(CD+MD)

=2AM^2+BM(BD+MD)+CM(CD-MD) ∵AM=CM=1/2BC

=2AM^2+BM(BD+MD+CD-MD)

=2AM^2+BM*BC

=2AM^2+2BM^2

=2(AM^2+BM^2)