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已知,如图,P是△ABC的内心,过点P作△ABC的外接圆的弦AE,交BC于点D,求证:BE=PE.
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人气:306 ℃ 时间:2019-08-29 04:36:45
解答
P是内心的作用:AE是∠BAC的平分线.
∵P在ΔABC的内心上,∴AE平分∠BAC,PB平分∠ABC,
∴∠EPB=∠PAB+∠PBA=1/2(∠BAC+∠ABC)
∵∠CBE=∠CAE=1/2∠BAC,
∴∠EBP=∠PBC+∠CBE=1/2(∠ABC+∠BAC)
∴∠EPB=∠EBP,
∴BE=PE.
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