Question

已知△ABC中,∠ABC,∠ACB的平分线交于E点.求证:∠BEC=90°+二分之一∠A

Answer 1

∠BEC = 180°- (∠EBC + ∠ECB)

2∠BEC = 360°- 2( ∠EBC + ∠ECB )

= 360°- ( 2∠EBC + 2∠ECB )

= 360°- ( ∠B + ∠C )

= 360°- ( 180°- ∠A )

= 180°+ ∠A

∠BEC=90°+二分之一∠A