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Browns’ Myles Garrett named AFC Defensive Player of the Week

Garrett, the No. 1 overall pick in 2017, had 4 1⁄2 sacks, a Browns record, on Sunday against the Bears at FirstEnergy Stadium.

Browns edge rusher Myles Garrett celebrates on Sunday.
Browns edge rusher Myles Garrett celebrates on Sunday.
Emilee Chinn/Getty Images

In one of the least surprising developments of the NFL season, Browns edge rusher Myles Garrett was named AFC Defensive Player of the Week on Wednesday morning.

Garrett, a former No. 1 overall pick, sacked the Bears 4 12 times, a Browns franchise record, on Sunday at FirstEnergy Stadium.

“I think they did a good job of moving him around a little bit,” Bears coach Matt Nagy said Monday. “They put him on the inside and did some things there. Really, no matter where he’s at whether he’s on the edge or whether he’s in the three-tech or shade [technique], he’s going to always be dominant. He’s a special player in this league. We knew that.

“You’ve got to always give respect to those type of players. And so now, we want to look into was it a time we got beat one on one, was it a time we got beat 2-on-1? What was the role of certain players in that specific call? We got to make sure we’re doing everything we can to make sure those players in the best situation possible.”

Bears offensive line coach Juan Castillo said the Bears tried — and failed — to not let Garrett beat them.

“That’s why you try to take him out of the game as best you can, you know?” he said.

Garrett said after the game it was easy to stop the Bears’ scheme, which did not feature as many rollouts of rookie quarterback Justin Fields as they’d expected.

“After that, I think we kind of settled in and saw how they planned to use the flow of the game,” Garrett said. “It kind of came to us easily after the second possession and [we] kind of figured out what they were going to do and how we were going to adjust to that.”