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20 lines
651 B
20 lines
651 B
2 years ago
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- 延长$AC$至$D$,使得$AD=AB$
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$\triangle ABP \cong \triangle ADP$
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也就是$\triangle APB$根据$AP$翻折后得到$\triangle APD$
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$\because \angle APB=150^{\circ}$
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$\therefore \angle BPD=60^{\circ}$
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并且$BP=PD$
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$\therefore \triangle PBD$是等边三角形
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$\because \angle PBC=30^{\circ}$
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$\therefore \angle CBD=30^{\circ}$
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$\triangle PBC \cong \triangle BCD$,$SAS$
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$\therefore PC=CD$
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$\therefore \angle CPD=\angle CDP$
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而$\angle PDC=\angle ABP=6^{\circ}$
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$\therefore \angle ACP=2*6^{\circ}=12^{\circ}$
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