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Last edited by hbghlyj 2022-3-11 12:06
M为$\overparen{BC}$中点,过C作BC垂线交BM于D,AD交BC于E,求证$2BC=AB+AC$⇔$2AE=3DE$.
证明:
$2AE=3DE\iff2\S{ABC}=3\S{BCD}\iff2Δ=\frac32a^2\tan\frac A2$
($∵\tan\frac A2=\frac{\sin A}{1+\cos A}$)$\iff4Δ=3a^2\frac{\sin A}{1+\cos A}\iff 2bc\sin A=3a^2\frac{\sin A}{1+\cos A}\iff2bc(1+\cos A)=3a^2$
$\iff 2bc+b^2+c^2-a^2=3a^2\iff b+c=2a$ |
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