Myles Garrett named AFC Defensive Player of the Week


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

BEREA, Ohio (WJW) – Defense end Myles Garrett has been named the AFC Defensive Player of the Week.

The NFL announced Wednesday the Browns star won the award.

Garrett set a Browns single-game record with 4.5 sacks.

The team handed a big loss to the Bears on Sunday 26-6.

Overall, the Browns’ defense finished with 9 sacks and held the Bears to 1-for-11 on third downs.

Garrett previously won the award in Week 4 last season.

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