Bills rookie DE Greg Rousseau named AFC Defensive Player of the Week

Buffalo Bills

KANSAS CITY, MISSOURI – OCTOBER 10: Darrel Williams #31 of the Kansas City Chiefs rushes the ball past the defense of Greg Rousseau #50 of the Buffalo Bills during the first half of a game at Arrowhead Stadium on October 10, 2021 in Kansas City, Missouri. (Photo by Jamie Squire/Getty Images)

ORCHARD PARK, N.Y. (WROC) — Buffalo Bills rookie defensive end Greg Rousseau has earned AFC Defensive Player of the Week honors following his impressive effort against the Kansas City Chiefs Sunday night.

Rousseaa registered a sack, five tackles, and a pivotal interception on a first and goal in the third quarter. He was a pivotal piece in the Bills’ dominant effort against the conference champions.

The rookie currently leads the Bills with three sacks through five games and joins quarterback Josh Allen (week 3) and linebacker Tremaine Edmunds (week 4) in winning conference player of the week honors.

This marks the first time that Buffalo has won the AFC Defensive Player of the Week award in back-to-back weeks since 2016.

The Bills currently lead the AFC East with a 4-1 record and have solidified their stance as one of the league’s best teams so far this season.

Buffalo travels to Tennessee to take on the Titans this week for Monday Night Football.

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