Cleveland Browns Star Earns Weekly NFL Honor

Myles Garrett won his fourth career weekly league award after three sack performance in the Browns win over the Steelers

Nov 21, 2024; Cleveland, Ohio, USA; Cleveland Browns defensive end Myles Garrett (95) celebrates a team fumble recovery against the Pittsburgh Steelers during the second quarter at Huntington Bank Field Stadium. Mandatory Credit: Scott Galvin-Imagn Images / Scott Galvin-Imagn Images

Cleveland Browns star Myles Garrett earned the AFC's Defensive Player of the Week following a three sack performance against the Pittsburgh Steelers.

All three sacks came in the first half for Garrett, who stole the show on Thursday Night Football, upstaging his AFC North rival T.J. Watt. Those efforts helped Garrett ascend the NFL's sack leaderboard as he now sits tied with Denver Broncos LB Nik Bonitto for the third overall in sacks this season. Only Danielle Hunter (10.5) of the Texans and Trey Hendrickson of the Bengals (11.5).

Garrett registered a forced fumble on one of his three sacks, late in the first half, which sparked a Browns scoring drive. Cleveland kicked a field goal just before the break to take a 10-3 lead into the half. The reigning Defensive Player of the Year had five tackles on the night as well.

Afterwards, he proclaimed that his performance was a statement that he is the best pass rusher in the league after Watt was held without a sack during the Browns victory.

Thursday's effort was Garrett's second three sack-game of the season, making him the only NFL player to record multiple games with at least three sacks in 2024.

Garrett has won four weekly awards throughout his eight NFL seasons. His first such honor came in Week 4 of the 2020 season, his second came one season later in Week 3 and he also won one in Week 7 of 2023. This marks the second time this season that a Browns player earned a weekly award after Jameis Winston earned one after a leading the Browns to a Week 8 victory over the Ravens.