题目内容
如图,在△ABC中,∠B=40°,△ABC的两个外角的平分线交于E点,求∠AEC的度数.
∵AE,CE是△ABC的两个外角的平分线,
∴∠ACE=
(∠B+∠BAC),∠CAE=
(∠B+∠BCA),
∵∠BCA+∠BAC=180°-∠B,
∴∠AEC=180°-∠ACE-∠CAE
=180°-
(∠B+∠BAC+∠B+∠BCA)
=180°-
(2∠B+180°-∠B)
=90°-
∠B.
=70°.
∴∠ACE=
1 |
2 |
1 |
2 |
∵∠BCA+∠BAC=180°-∠B,
∴∠AEC=180°-∠ACE-∠CAE
=180°-
1 |
2 |
=180°-
1 |
2 |
=90°-
1 |
2 |
=70°.
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