Question

如图,在△ABC中,BE平分∠ABC交AC于E,CD⊥AC交AB于D,∠BCD=∠A.求∠BEA的度数.

Answer 1

在△ABC中

∠A+∠ACB+∠ABC=180°

而∠ACB=90°+∠DCB

∠DCB=∠A

∴∠A+90°+∠A+∠ABC=180°

2∠A+∠ABC=90°

∠A+1/2∠ABC=45°

∵BE平分∠ABC

∴∠ABE=1/2∠ABC

∴∠A+∠ABE=45°

∴∠BEA=135°