题目如图,DE⊥ABDE\bot ABDE⊥AB于EEE,DF⊥ACDF\bot ACDF⊥AC于FFF,若BD=CDBD=CDBD=CD,ADADAD平分∠BAC\angle BAC∠BAC,则下列结论错误的是( )A.DE=DFDE=DFDE=DFB.BE=CFBE=CFBE=CFC.∠ABD+∠C=180∘\angle ABD+\angle C=180^{\circ}∠ABD+∠C=180∘D.AB+AC=2ADAB+AC=2ADAB+AC=2AD知识点:展开图折叠成几何体、角平分线的性质章节:未标注