Buffalo Bills cornerback Rasul Douglas has been named the AFC Defensive Player of the Week for his performance in the Bills' Week 17 win over the New England Patriots.
Douglas had three pass breakups – including one which allowed defensive tackle Ed Oliver a chance to make an interception – and two interceptions of his own, one of which he returned 40 yards for a touchdown.
Douglas, who the Bills acquired at the trade deadline, has been an integral part of the Buffalo secondary since making his first appearance in blue and red in Week 9 against the Cincinnati Bengals. The eight-year pro has been on the field for every defensive snap in five of his seven active games and has created splash plays throughout his time in Buffalo.
"I feel like he was everywhere," said defensive tackle DaQuan Jones after the Patriots game. "Like, every time the ball was thrown, he was right there. I mean, he was a special player, I'm so happy we got him over here."
Across the eight games Douglas has played in Buffalo, he has accounted for four interceptions, two fumble recoveries, eight pass breakups and a sack. Douglas said that his aggressive mentality has led to him becoming such an explosive player.
"If you make the play, ain't nobody gonna say nothing," Douglas said. "It's like Steph Curry, like he can shoot all the crazy shots he wants, but if he's making them, what are you gonna say as a coach? … [defensive backs coach John Butler], he allows me to play. He trusts that I watch film and the film he gives me to watch."
This marks Douglas' second Player of the Week award, as he achieved the honor in Green Bay's Week 12 win over the Los Angeles Rams after tallying four pass breakups and a pick-six.