选B
EH = FG = BD/2, EF = HG = AC/2
∵EH∥FG
∴∠FEH + ∠EFG = 180°
∴cos∠FEH + cos∠EFG = 0
根据余弦定理：
在△EFH中,EH² + EF² - 2EH·EF·cos∠FEH = FH² = 16 ……①
在△EFG中,FG² + EF² - 2EF·FG·cos∠EFG = EG² = 9
∵EH = FG
∴EH² + EF² - 2EF·EH·cos∠EFG = EG² = 9……②
① ＋ ②,得
2(EH² + EF²) =25
∴AC²+BD² = (2EF)² + (2EH)² = 4(EH² + EF²)=50