Question

如图①,△ABC中,∠ABC的外角平分线与∠ACB的外角平分线交于D(1)∠D=50°,求∠A的度数;(2)探究

如图①,△ABC中,∠ABC的外角平分线与∠ACB的外角平分线交于D(1)∠D=50°，求∠A的度数;(2)探究∠A与∠D之间的数量关系；(3)如图②,连AD,求证:AD平分∠BAC.

Answer 1

1
∠DCE=∠DBE+∠D
∠D=∠DCE-∠DBE
=∠ACE/2-∠ABE/2
=(1/2)(∠A+∠ABE)-∠ABE/2
=(1/2)∠A+∠ABE/2-∠ABE/2
=(1/2)∠A

2
不成立；
∠D=180°-∠DBC-∠DCB
=180°-(1/2)∠EBC-(1/2)∠FCB
=180°-(1/2)(∠A+∠ACB)-(1/2)(∠A+∠ABC)
=180°-(1/2)∠A-(1/2)∠ACB-(1/2)∠A-(1/2)∠ABC)
=180°-∠A-(1/2)(∠ACB+∠ABC)
=180°-∠A-(1/2)(180°-∠A)
=180°-∠A-(1/2)180°+(1/2)∠A
=90°-(1/2)∠A

Answer 2

1，角A=80度
2，角A=180度—2×角D
3，回家再做

Answer 3

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