AE²+PE²=AP²=AF²+PF²
BD²+PD²=BP²=BE²+PE²
CF²+PF²=CP²=CD²+PD²
三式AE²+PE²=AF²+PF²,BD²+PD²=BE²+PE²,CF²+PF²=CD²+PD²相加
得AE²+PE²+BD²+PD²+CF²+PF²=AF²+PF²+BE²+PE²+CD²+PD²
整理得AE²+BD²+CF²=AF²+BE²+CD²
周长为a,等边三角形边长为a/3
由
AF=a/3-CF
BE=a/3-AE
CD=a/3-BD
代入式子AE²+BD²+CF²=AF²+BE²+CD²
AE²+BD²+CF²=(a/3-CF)²+(a/3-AE)²+(a/3-BD)²
整理
AE²+BD²+CF²=a²/9-(2a/3)CF+CF²+a²/9-(2a/3)AE+AE²+a²/9-(2a/3)BD+BD²
0=a²/3-(2a/3)CF-(2a/3)AE-(2a/3)BD
(2a/3)(AE+BD+CF)=a²/3
AE+BD+CF=(a²/3)(3/2a)=a/2