回復 33# nanpolend 的帖子
取BC中點D，那麼
PB2+PC2=2(PD2+BD2)...........(1)
GB2+GC2=2(GD2+BD2)...........(2)
取AG中點K，那麼AK=KG=GD
PA2+PG2=2(PK2+AK2)............(3)
PK2+PD2=2(PG2+KG2)............(4)
(1)-(2)得
PB2+PC2+2GD2=GB2+GC2+2PD2)...............(5)
(3)+(4)*2得
PA2+2PD2=2AK2+3PG2+4KG2=6AK2+3PG2...........(6)
(5)+(6)得
PA2+PB2+PC2=GB2+GC2+4AK2+3PG2=GA2+GB2+GC2+3GP2
