Myles Garrett named AFC Defensive Player of the Week following win vs. Bears

Browns DE Myles Garrett was named the AFC Defensive Player of the Week after his 4.5-sack game vs. Justin Fields and the Bears.

With Week 3 in the books, the NFL announced its players of the week among the NFC and AFC. And with the way the Cleveland Browns defense delivered a beatdown of the Chicago Bears offense on Sunday, it’s not a surprise to find defensive end Myles Garrett among those honored.

Garrett was named the AFC Defensive Player of the Week for his 4.5-sack performance against rookie quarterback Justin Fields. Garrett had 7 tackles, including 4 tackles for loss, as well as 6 QB hits to add to his sack total. The 4.5 sacks set a Browns record. In all, Cleveland sacked Fields nine times on the afternoon.

The Bears offense was limited to just 47 — that’s right, 47 — yards of offense thanks to Matt Nagy’s atrocious game plan and an offensive line that didn’t stand a chance against Garrett and the Browns’ defensive front.

This is the second time this season that an opposing player has been named a player of the week. Rams quarterback Matthew Stafford was named NFC Offensive Player of the Week for his near-flawless performance against the Bears on Sunday Night Football in Week 1.

Chicago will be looking to rebound following another brutal loss when they host the Detroit Lions on Sunday.

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