AM為三角形ABC中BC上的中點求證：AB*AB+AC*AC=AM*AM+BM*BM

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AM為三角形ABC中BC上的中線求證：AB*AB+AC*AC=2(AM*AM+BM*BM)過A作AN垂直BC于N，則AB^2 = AN^2 + BN^2 ,AC^2 = AN^2 + CN^2所以　AB^2 ＋ AC^2 = AN^2 + BN^2 + AN^2 + CN^2= 2AN^2 + (BM+MN)^2 + (BM-MN)^2= 2AN^2 + 2 BM^2 + 2MN^2= 2BM^2 + 2(AN^2 + MN^2)= 2BM^2 + 2AM^2