证明:过A做AE⊥BC于E,如图
∵AM是△ABC的中线
∴MC=BM
∴AB²+AC²
=AE²+BE²+AE²+EC²
=2AE²+(BM+EM)²+(MC-EM)²
=2AE²+(BM+EM)²+(BM-EM)²
=2(AE²+EM²)+2BM²
=2(AM²+BM²)
即AB²+AC²=2(AM²+BM²)
证明:过A做AE⊥BC于E,如图
∵AM是△ABC的中线
∴MC=BM
∴AB²+AC²
=AE²+BE²+AE²+EC²
=2AE²+(BM+EM)²+(MC-EM)²
=2AE²+(BM+EM)²+(BM-EM)²
=2(AE²+EM²)+2BM²
=2(AM²+BM²)
即AB²+AC²=2(AM²+BM²)