Myles Garrett named AFC Defensive Player of the Week

Chicago Bears v Cleveland Browns

CLEVELAND, OHIO - SEPTEMBER 26: Myles Garrett #95 of the Cleveland Browns celebrates a sack during the second quarter in the game against the Chicago Bears at FirstEnergy Stadium on September 26, 2021 in Cleveland, Ohio. (Photo by Emilee Chinn/Getty Images)Photo: (Photo by Emilee Chinn/Getty Images)

BEREA, Ohio – Cleveland Browns DE Myles Garrett has earned AFC Defensive Player of the Week for games played Sept. 23-27 (Week 3), the National Football League announced Wednesday.

Garrett set a Browns single game record with 4.5 sacks during the team’s 26-6 victory over the Bears on Sunday. He added seven tackles and four TFLs. Garrett helped the Browns establish single-game team records for net yards allowed (47) and net passing yards allowed (1). The Browns defense finished with nine sacks and held the Bears to 1-for-11 on third downs.

This marks Garrett’s second career Player of the Week award as he also earned the honor in Week 4 last season.

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