Greg Rousseau named AFC Defensive Player of the Week

This marks the second time that Rousseau has been honored as the AFC Defensive Player of the Week in his career

Orchard Park N.Y. (WGR Sports Radio 550) - Following an impressive Week 1 performance on the defensive side of the ball against the Arizona Cardinals, Buffalo Bills defensive end Greg Rousseau has earned honors as the AFC Defensive Player of the Week.

Rousseau started hot out of the gate with a six-tackle performance, which included a career-best three sacks of Cardinals quarterback Kyler Murray. One of those sacks also resulted in a forced turnover in the third quarter of Buffalo's 34-28 win at Highmark Stadium.

With his three sacks on Sunday, Rousseau tied a franchise-best mark for the most sacks in a season opener with Ben Williams (1983).

This marks the second time in Rousseau's career that he earns the honor as the AFC Defensive Player of the Week. He also earned the same honor during his rookie season in 2021 during a Week 5 win over the Kansas City Chiefs.

The Bills will next play the Miami Dolphins this Thursday night in a huge early season divisional matchup at Hard Rock Stadium in South Florida.

Kickoff is slated for 8:15 p.m. ET on WGR with Chris Brown, Eric Wood, and Sal Capaccio on the call.

Photo credit Losi & Gangi
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