如图所示，已知在四边形ABCD中，BC﹥AB,AD=CD,BD平分∠ABC.试说明∠BAD+∠C=180°
有2个网友回答
解：在BC上取一点E，使BE=BA，连结ED
∵BD平分∠ABC
∴∠ABD=∠CBD
又∵BE=BA,BD=BD
∴△ABD≌△EBD（SAS）
∴∠A=∠BED，AD=ED
又∵AD=CD
∴ED=CD
∴∠C=∠DEC
∠A+∠C=∠BED+∠DEC=180°
证明：过点D作DE⊥AB交BA的延长线于点E，DF⊥BC于F
∵BD平分∠ABC，DE⊥AB，DF⊥BC
∴DE＝DF（角平分线性质），∠DEA＝∠DFC＝90
∵AD＝CD
∴△ADE≌△CDF （HL）
∴∠DAE＝∠C
∵∠BAD+∠DAE＝180
∴∠BAD+∠C＝180
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