Question

如图，已知DE”BC，CD是角ACB的平分线，角B=70度，角ACB=50度，求角EDC和角BDC

如图，已知DE”BC，CD是角ACB的平分线，角B=70度，角ACB=50度，求角EDC和角BDC的度数

Answer 1

∵DE//BC
∴∠EDC=∠BCD
∵DC平分∠ACB
∴∠DCB=∠ACB/2
∵∠ACB=50°
∴∠DCB=25°
∴∠EDC=25°
∵∠BDC+∠B+∠DCB=180°
∠B=70° ∠DCB=25°
∴∠BDC=85°

Answer 2

没图吗