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如图,在等腰△ABC中,AB+=AC,点D在AC边上,连接BD,已+知BD+=AB,∠CBD=18°,求∠
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咨询记录 · 回答于2023-06-28
如图,在等腰△ABC中,AB+=AC,点D在AC边上,连接BD,已+知BD+=AB,∠CBD=18°,求∠
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,很高兴为您解答:根据题意可得:AB = AC (等腰三角形两边相等)BD = AB (已知BD = AB)∠CBD = 18°由三角形内角和公式可得:∠ABC = ∠ACB = (180° - ∠BAC) / 2因为 AB = AC,所以 ∠ABC = ∠ACB又因为 BD = AB,所以 ∠BDC = ∠BCD因此,∠CBD = ∠ACB - ∠BCD代入已知条件可得:18° = (180° - ∠BAC) / 2 - (180° - ∠BAC) / 4化简可得:∠BAC = 72°因为 AB = AC,所以 ∠ABC = ∠ACB = (180° - ∠BAC) / 2 = 54°因为 BD = AB,所以 ∠BDC = ∠BCD = (180° - ∠CBD) / 2 = 81°因此,答案为∠ABC = ∠ACB = 54°。
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