如图,已知三角形ABC中,点M为AB的中点,并且CM=二分之一AB,求证BC垂直AC

1、∵点M为AB的中点,CM=1/2AB∴AM=BE=1/2AB=CM∵AM=CM∴∠A=∠MCA∵BM=CM∴∠B=∠MCB∴∠A+∠B=∠MCA+∠MCB∵∠ACB=∠MCA+∠MCB∴∠A+∠B=∠ACB∵∠A+∠B+∠ACB=180°∴∠ACB=90°即BC⊥AC2、设5x²-2xy-y²=0△=(2y)²-4×5×(-y²)=24y²=2²×6×y²∴x=(2±2√6)y/10=(1±√6)/5∴原式=[x-(1+√6)/5][x-(1-√6)/5]=1/25(5x-y-√6y)(5x-y+√6y)