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