Question

DE分别与△ABC的边AB、BC相交于D、E 与BC的延长线相交于F,∠B＝65°∠ACB＝80°∠AED＝50°求∠B

Answer 1

DE分别与△ABC的边AB、AC相交于D、E 与BC的延长线相交于F,∠B＝65°∠ACB＝80°∠AED＝50°求∠BDF的度数

∠AED＝50°=∠CEF

∠F=∠ACB-∠CEF=80°-50°=30°

∠BDF=180°-30°-65°=85°