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BEREA, Ohio – Cleveland Browns DE Myles Garrett has earned AFC Defensive Player of the Week for games played Oct. 1-5 (Week 4), the National Football League announced Wednesday.

During a 49-38 win at Dallas on Sunday, Garrett recorded three tackles, two sacks and one forced fumble. He helped the Browns improve to 3-1, their best start to a season since 2001. Garrett has recorded at least one sack and one forced fumble in three consecutive games.

The Browns are leading the NFL with 10 takeaways and a +6 turnover margin this season. Garrett is tied for the NFL lead in sacks (five), forced fumbles (three) and fumble recoveries (two).

Garrett is the first Browns player to win AFC Defensive Player of the Week since LB Joe Schobert in Week 12 last year and the first Browns defensive end to win the award since Kenard Lang in Week 1 of 2004. This is the first league award in Garrett’s career.

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