For the third time in four years, Aaron Donald has been named defensive player of the year by the Associated Press.
Donald is the third three-time winner of the award, joining Hall of Fame linebacker Lawrence Taylor and Texans defensive lineman J.J. Watt. The honor was given out at the NFL Honors show in Los Angeles on Saturday.
“To be the third to do it that is truly a blessing. It shows the body of work that I have; anytime your hard work is rewarded you are going to be happy about that. It is just a blessing,” Donald said, via the Associated Press.
Donald had 13.5 sacks in 2020, the second-highest total of his career. The defensive tackle also had 14 tackles for loss, four forced fumbles, a fumble recovery, and 28 quarterback hits.
The Rams’ defensive tackled edged out Steelers linebacker T.J. Watt for the award, 27-20. Watt led the league with 15.0 sacks and 23 tackles for loss, also registering seven passes defensed, two forced fumbles, and 41 quarterback hits.
Dolphins cornerback Xavien Howard, who led the league with 10 interceptions and 20 passes defensed, received the remaining three votes.
“From stopping the run to rushing the passer to just trying to dominate, I think I am a rounded player,” Donald said. “I feel I can do everything. No matter the size of the person, as long as you can play at high level and be productive, that is all that matters.
“I will always say my overall game, my mindset is always trying to find ways to get better, stay consistent from stopping the run and rushing the passer. When you make a name for yourself it just gets 10 times harder. So I got to always find ways to improve as a rounded football player, and that’s what I will continue to do.”
Donald previously won the award in 2017 and 2018.