Question

如图,已知○O内接正十边形ABCD···AD交OB、OC于M、N求证 MN∥BC MN+CO=OB

Answer 1

(1)∵ABCDEFGHIJ是正十边形,∴AB=BC=CD;

∵内接于圆O;OB=OC=R;

弧AC=2弧BC;

∴弧AC对应的圆周角∠ADC=弧BC对应的圆心角∠BOC;

∠BCD=∠OCB+∠OCD;

易证:∠ADC+∠OCB+∠OCD=180度;既∠ADC+∠BCD=180度;

∴BC//AD;即BC//MN;

(2)连接OA AB

∵∠BOC=360度/10=36度;

∵∠ONA=∠OCB=(180-36)/2=72度;

∵∠AOC=2*BOC=2*36=72度;

∴∠ONA=∠AOC;

∴OA=AN=AM+MN;

∵∠ABM=∠AMB=72度;

∴AM=AB=BC;

∴OB=OA=AM+MN=BC+MN;

∴MN+BC=OB;