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NFL announces Browns DE Myles Garrett as AFC Defensive Player of the week

This comes after Garrett broke his own single-season career sack record in the Browns Week 12 game against the Las Vegas Raiders
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Cleveland Browns defensive end Myles Garrett was named AFC Defensive Player of the week on Wednesday by the National Football League.

This comes after Garrett broke his own single-season career sack record in the Browns Week 12 game against the Las Vegas Raiders— setting a new franchise record of 18.

In the 24-10 win, the defensive end made five tackles, four tackles for loss, three sacks and two forced fumbles.

Garrett led all league players in sacks and forced fumbles for Week 12.

Cleveland's defensive had 10 sacks, which is the most of any NFL team this season and second-most by a Browns team in a game.

Garrett leads the NFL this season in sacks (18) and tackles for loss (26).

This is his fifth career weekly league award and fifth since Week 12 in 2024. He joins kicker Andre Szmyt, who won AFC Special Teams Player of the Week for Week 3, and safety Grant Delpit, who won AFC Special Teams Player of the Week during Week 7, as the third Browns player to be named AFC Player of the Week this season.

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