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Last edited by hbghlyj at 2024-12-13 17:42:0056楼
以$B$为圆心,$AB$为半径作圆交$BC$于$D$
以$C$为圆心,$AC$为半径作圆交$BC$于$E$
两圆交于$A,F$,
$FD$交圆$C$于$D,G$,
$FE$交圆$B$于$F,H$,
$DH,EG$交于$I$,
求证$I$是$△ABC$的内心.
证明
2∠FHD=∠FBC=∠FBA/2=∠FHA⇒HI平分∠AHE.
∠AHI=∠AFD=∠AEI⇒AHEI共圆.
∴AI=IE,又AC=CE,∴CI平分∠ACB,同理BI平分∠ABC. |
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