In a standout performance during Week 8 of the NFL season, Tee Higgins of the Cincinnati Bengals has been named the AFC Offensive Player of the Week. The announcement was made by the NFL on Wednesday, recognizing Higgins for his impressive contributions in a thrilling 43-36 overtime victory against the San Francisco 49ers.

Higgins recorded 10 receptions for 155 yards and scored two touchdowns, playing a crucial role in the Bengals' comeback. His first touchdown helped the team take a 21-13 lead in the third quarter, while his second touchdown was pivotal in tying the game, showcasing his ability to perform under pressure. This performance edged out his teammate Ja'Marr Chase, who also had a notable game with 11 catches for 141 yards and one touchdown.S1S2

Higgins's recognition as the AFC Offensive Player of the Week highlights not only his individual talent but also the depth of the Bengals' receiving corps, which has been a significant factor in their offensive strategy this season.S1