Midgarding
幼苗
共回答了20个问题采纳率:95% 举报
∠AED + ∠BDE = ∠AEC + ∠CED + ∠BDC + ∠CDE
= 90° + (180° - ∠ECD)
=∠ACB
过C作∠ACF=∠AED使CF = ME = MD
则∠BCF = ∠BDE
连AF,BF,MF
则△ACF≌△AEM,△BCF≌△BDM
∴AM=AF,∠CAF=∠EAM
∴∠MAF=∠EAC=90°
∴∠AMF=45°
同理∠BMF=45°
∴∠AMB = ∠AMF+∠BMF = 90°
即AM⊥BM
1年前
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