回復 33# nanpolend 的帖子
取BC中點D，那麼
\(\displaystyle PB^2+PC^2=2(PD^2+BD^2) \)...........(1)
\(\displaystyle GB^2+GC^2=2(GD^2+BD^2) \)...........(2)
取AG中點K，那麼AK=KG=GD
\(\displaystyle PA^2+PG^2=2(PK^2+AK^2) \)............(3)
\(\displaystyle PK^2+PD^2=2(PG^2+KG^2) \)............(4)
(1)-(2)得
\(\displaystyle PB^2+PC^2+2GD^2=GB^2+GC^2+2PD^2) \)...............(5)
(3)+(4)*2得
\(\displaystyle PA^2+2PD^2=2AK^2+3PG^2+4KG^2=6AK^2+3PG^2 \)...........(6)
(5)+(6)得
\(\displaystyle PA^2+PB^2+PC^2=GB^2+GC^2+4AK^2+3PG^2=GA^2+GB^2+GC^2+3GP^2 \)
