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.亲们帮帮忙,解答一下~~

Question

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.
亲们帮帮忙,解答一下~~

Answer 1

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.