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