Question

几何证明题

求证：四边形各边平方的和等于对角线平方的和，加上对角线中点连线平方的四倍。

Answer 1

解：根据余弦定理，AD²=AG²+DG²-2AG*DG*C0Sα,BC²=BG²+CG²-2BG*CG*C0Sα

  AB²=AG²+BG²-2AG*BG*C0Sβ,  DC²=DG²+CG²-2DG*CG*C0Sβ,

  EF²=EG²+FG²-2EG*FG*C0Sβ

   ∵α+β=180° ∴C0Sβ=-C0Sα

   ∵AE=EC  ∴EG=(AG-CG)/2,  同理FG=（BG-DG）/2

   ∴4EF²=4*(EG²+FG²-2EG*FG*C0Sβ)=(AG-GC)²+（BG-DG)²-2(AG-CG)*(BG-DG)*C0Sβ

         =AG²+BG²+CG²+DG²-2*AG*CG-2*BG*DG-2(AG*BG-AG*DG-CG*BG+CG*DG)*C0Sβ

         =(AG²+BG²-2AG*BG*C0Sβ)+(CG²+DG²-2CG*DG*C0Sβ)-2*AG*CG-2*BG*DG+2(AG*DG+CG*BG)*C0Sβ

        =AB²+DC²-2*AG*CG-2*BG*DG-2*AG*DG*C0Sα-2*CG*BG*C0Sα

        =AB²+DC²-2*AG*CG-2*BG*DG+(AD²-AG²-DG²)+(BC²-BG²-CG²)

        =AD²+BC²+AB²+DC²-(AG²+CG²+2*AG*CG)-(BG²+DG²+2*BG*DG)

        =AD²+BC²+AB²+DC²-(AG+CG)²-(BG+DG)²

        =AD²+BC²+AB²+DC²-AC²-BD²

     ∴4EF²=AD²+BC²+AB²+DC²-AC²-BD²

         AD²+BC²+AB²+DC²=AC²+BD²+4EF²,得证。