Question

如图,在直角三角形ABC中,∠ACB=90度,∠A=22.5度,AB的中垂线DE交BC于点E,F为AC上一点,AG⊥EF

Answer 1

∠A=22.5度打错,是∠B=22.5度

①⊿ACH∽⊿EGH.（∠ACH＝∠EGH＝90º,∠H公共）,∴CH/CA=CF/CE

②连接AE.∴DE是中垂线.∠EAD＝∠B＝22.5º,∴∠AEC＝2×22.5º＝45º

⊿AEC等腰直角,CA＝CE.从CH/CA=CF/CE.即有CH=CF