题目内容
已知:如图,AB∥DE,∠A=∠D,且BE=CF,
求证:∠ACB=∠F.
∵AB∥DE,∴∠B=∠DEF,·························· 1分
∵BE=CF, ∴BE+CE=CF+CE,即BC=EF,············· 2分
∵∠A=∠D,∴△ABC≌△DEF.·················· 4分
∴∠ACB=∠F.
解析:略
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题目内容
已知:如图,AB∥DE,∠A=∠D,且BE=CF,
求证:∠ACB=∠F.
∵AB∥DE,∴∠B=∠DEF,·························· 1分
∵BE=CF, ∴BE+CE=CF+CE,即BC=EF,············· 2分
∵∠A=∠D,∴△ABC≌△DEF.·················· 4分
∴∠ACB=∠F.
解析:略