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Chiefs' LB Dee Ford Earns AFC Defensive Player of the Week Honors

Ford tallied three sacks and two forced fumbles in Sunday’s victory over Denver

Kansas City Chiefs' linebacker Dee Ford was creating havoc defensively throughout Sunday's victory over the Denver Broncos, putting together a performance that caught the eye of those around the league.  

The NFL named Ford the AFC Defensive Player of the Week on Wednesday, marking the first time that the fifth-year linebacker has won the award.

Ford recorded four tackles, three sacks and two forced fumbles on Sunday, becoming just the fifth player in the last three seasons to tally at least three sacks and two forced fumbles in a single game.

He also notched eight quarterback hurries on the afternoon, according to Pro Football Focus, and now has a league-leading 45 through eight games.

Ford, who is the first Chiefs' linebacker to win the award since Tamba Hali in 2014, is now just two sacks shy of matching his career-high (10), and there's still eight games left in the campaign.

He'll try to get to the quarterback yet again this weekend as the Chiefs take on the Cleveland Browns.

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