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。
