New England Patriots linebacker Tedy Bruschi has been named the AFC Defensive Player of the Week for his performance against the Buffalo Bills last Sunday.
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Bruschi chalked-up eight tackles, including 2 sacks, during New England's 31-17 win over the Bills.
Leading 24-17 the veteran linebacker forced a game-clinching Buffalo fumble on 4th-and-3 at the New England 17-yard line with 2:59 left in the game, which teammate Richard Seymour scooped up and ran back 68 yards for a touchdown. The play sealed the victory for the Patriots who have won 18 consecutive games.
This is the third Player of the Week distinction Bruschi has received, all of which have been awarded since Week 2 of last season.