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Cleveland Browns DE Myles Garrett named AFC Defensive Player of the Week after win over Pittsburgh Steelers

The reigning NFL Defensive Player of the Year's three sacks and one forced fumble played a crucial role in Cleveland's shocking upset over the AFC North leaders.

CLEVELAND — After a monster performance in a thrilling home victory against fierce rivals the Pittsburgh Steelers, Cleveland Browns defensive end Myles Garrett declared, "I'm the guy." Now, he has even more proof. 

The reigning NFL Defensive Player of the Year recorded three sacks and a forced fumble to lift the Browns (3-8) over the Steelers (8-3), a yeoman's effort that has earned him AFC Defensive Player of the Week honors for Week 12, the NFL announced Wednesday. 

RELATED: Cleveland Browns' Myles Garrett gets 3 sacks in win, sends message to Pittsburgh Steelers and T.J. Watt: 'I'm the guy'

Garrett's three sacks each played a major role in Cleveland's shocking upset over the AFC North leaders. The first came on a third down in Cleveland territory that forced a longer, and ultimately unsuccessful, Pittsburgh field goal attempt. The second forced a fumble that the Browns recovered and converted into points via a field goal. The third time Garrett dumped quarterback Russell Wilson in the backfield ended the first half on a third-down play near midfield.

Garrett's 10 total sacks this year are good for third in the league. He is the only player in the NFL who has recorded multiple games with at least three sacks this season and the only NFL player with at least 10 sacks in each of the past seven seasons dating back to 2018. 

This is Garrett's fourth career Player of the Week nod. Quarterback Jameis Winston, the Offensive Player of the Week for Week 8, is the only other Brown to have earned the honor this year.

   

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