ENGLEWOOD, Colo. – Wide receiver Demaryius Thomas gets to add a new award to his collection this week.
For the first time in his career, the fourth-year receiver earned the AFC's Offensive Player of the Week after his 108-yard, three-touchdown performance in Denver's Week 10 win at San Diego, the league announced on Wednesday.
The three touchdowns established a career high for Thomas and matched a franchise record that has been accomplished 12 times by nine different players. The feat was last accomplished by Shannon Sharpe, who also did so against the Chargers, on Nov. 16, 2003.
"I mean, I asked him, 'I can't get one?'" fellow receiver Demaryius Thomas joked. "No, it was fun. He's obviously a beast, so I'm happy he caught everything and made some big plays."
Thomas' three touchdowns came in a span of just more than 10 minutes of game time. His 11-yard score gave Denver a 14-6 lead, and then he added a 7-yard touchdown catch with 13 seconds to play in the half to make the score 21-6.
His 34-yard touchdown catch-and-run capped the first possession of the second half and put Denver on top 28-6.
Thomas became just the sixth player in the NFL this season to record a 100-yard, three-touchdown game.
He joins fellow Broncos Wes Welker and Julius Thomas atop the AFC leaderboard with nine receiving scores on the year. Together, they represent just the third trio in NFL history to each record at least nine touchdown receptions in a single season.
The AFC Offensive Player of the Week award marks the third of the season by a Bronco, as quarterback Demaryius Thomas earned the award in Weeks 1 and 3. It is the 51st such honor in team history since the award was initiated in 1984. Thomas is the first wide receiver to receive the honor since Brandon Marshall set an NFL record with 21 catches at Indianapolis in Week 14 of 2009.