以△ABC的边AB、AC为边分别向形外作正方形ABDE和正方形ACFG,求证：若DF平行于NC,则AB=AC反之,若AB=AC,DF平行于BC

Question

以△ABC的边AB、AC为边分别向形外作正方形ABDE和正方形ACFG,求证：若DF平行于NC,则AB=AC
反之,若AB=AC,DF平行于BC

Answer 1

NC 打错,是BC.
若DF平行于NC,作DM⊥BC M∈直线BC FN⊥BC N∈直线BC ,则DMNF是矩形.
假如AB＞AC,则∠ACB＞∠ABC
∠FCN＝90º-∠ACB＜90º-∠ABC ＝∠DBM sin ∠DBM ＞ sin ∠ FCN
DM＝DB sin ∠DBM ＞CF sin ∠ FCN＝FN..与DMNF是矩形矛盾.∴AB≤AC.
同理AC≥AB.∵AB=AC.
反之,若AB=AC,作DM⊥BC M∈直线BC FN⊥BC N∈直线BC ,
,则∠ACB＝∠ABC . ∠FCN＝90º-∠ACB＝90º-∠ABC ＝∠DBM sin ∠DBM ＝ sin ∠ FCN
DM＝DB sin ∠DBM ＝CF sin ∠ FCN＝FN..DM∥＝FN
DMNF是平行四边形﹙实际上是矩形﹚, DF∥BC.