jf1199
幼苗
共回答了17个问题采纳率:88.2% 举报
因为∠BAC的外角平分线交CB的延长线于点E
所以 2∠EAB+∠BAC=180
∠EAB+∠ABE+∠E=180 ∠ACB+∠BAC+∠ABC=180
∠ABC=E+EAB ∠EAB=∠ABC-∠E
2∠EAB=∠ABC+∠ACB=180 2(∠ABC-∠E)=∠ABC+∠ACB
∠ABC-∠E=1/2 ∠ABC+∠ACB
-∠E=1/2∠ABC+∠ACB-∠ABC
∠E=∠ABC+1/2(∠ABC-∠ACB)
∠E=二分之一(∠ABC-∠ACB)
1年前
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