∴∠DBC=
1 |
2 |
1 |
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∵∠ABC+∠ACB=180°-∠A,
∠BDC=180°-∠DBC-∠DCB=180°-
1 |
2 |
1 |
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1 |
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∴∠BDC=90°+
1 |
2 |
(2)∵BD、CD是∠ABC和∠ACB外角的平分線,
∴∠CBD=
1 |
2 |
1 |
2 |
∵∠ABC+∠ACB=180°-∠A,
∠BDC=180°-∠CBD-∠BCD=180°-
1 |
2 |
=180°-
1 |
2 |
1 |
2 |
即∠BDC=90°-
1 |
2 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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