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[几何] 直角三角形求线段长度

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乌贼 posted 2021-9-17 14:35 |Read mode
Last edited by 乌贼 2021-9-18 01:33转自 tieba.baidu.com/p/7508701432
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如图: 211.png
易证$ \angle QPF=\angle QFP=45\du  $,$ ABPNM $五点共圆\[ BE=2EN\riff MN=3EN\riff BP=3PE\riff BC=3CE\riff BC=CE \]即$ BPNF $为平行四边形,$ BFP $为等腰直角三角形\[ BF=\dfrac{6\sqrt{5}}{5} \]

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isee posted 2021-9-18 00:27
回复 1# 乌贼

题也弄过来,哪怕是图也好。

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original poster 乌贼 posted 2021-9-18 01:33
回复 2# isee
弄过来了

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