∵AB=AC∴∠B=∠C又∠BAC+∠B+∠C=180°108°+2∠C=180°∠C=36° ∵∠B=∠C且AD⊥BC则AD为BC的中垂线则∠BAD=∠CAD又∠BAD+∠CAD=∠BAC=108°则∠BAD=1/2∠BAC=54° 满意请采纳,谢谢~
∠C=(180°-108°)÷2=36°∠BAD=108°÷2=54°
∠C=∠B=36°∠BAD=90°-36°=54°