Cho tam giác ABC có ba góc nhọn đường cao AH

a. CMR: AB² + AC² = \(\frac{{BC}}{2}\) + 2AM²

b. AC² − AB² = 2BC.HM (AC > AB)

a) Ta có: AB² + AC² = 2AH² + BH² + CH²

= 2AM² − 2HM² + (BM − HM)² + (CM + HM)²

= 2AM² − 2HM² + BM² − 2BM.HM + HM² + CM² + 2CM.HM + HM²

= 2AM² + BC² − 2BM.CM = 2AM² + BC² − \(\frac{{2B{C^2}}}{4}\)

= 2AM² + \(\frac{{B{C^2}}}{2}\)(đpcm)

b) Ta có: AC² – AB² = AH² + HC² – BH² – AH²

= HC² – BH² = (CM + HM)² − (BM − HM)²

= CM² + 2CM.HM + HM² – BM² + 2BM.HM – HM²

= 2HM(CM + BM)

= 2HM.BC (đpcm).