Former K-State basketball star Dean Wade agrees to 4-year deal with new NBA team
AI-generated summary reviewed by our newsroom.
- Dean Wade agreed to a four-year, $39 million contract with the Philadelphia 76ers.
- Wade spent seven seasons with the Cavaliers and is entering his eighth NBA year.
- Wade averaged 5.3 points and 3.6 rebounds per game during his NBA career to date.
Dean Wade is on the move.
The former Kansas State basketball star from St. John has ended his time as a free agent by agreeing to a new four-year contract with the Philadelphia 76ers that is reportedly worth $39 million.
Wade is entering his eighth year in the NBA, an impressing accomplishment when you consider he went undrafted in 2019 after a stellar four seasons with the Wildcats.
Back then, Wade signed a two-way contract with the Cleveland Cavaliers as an undrafted free agent. He went on to make the Cleveland roster that season and later become a mainstay in the rotation. Wade spent seven seasons with the Cavaliers, appearing in 342 games with 160 starts.
He has averaged 5.3 points and 3.6 rebounds as a professional. His teammates and coaches have regularly praised him as a valuable defender, shooter and role player in the NBA.
Wade, 29, is a 6-foot-9 and 228-pound forward who is ready to cash in with a new team as the 76ers will likely use him in similar ways.
Before he reached the NBA, Wade was an All-Big 12 player at K-State.
As a senior, Wade helped the Wildcats win a conference championship by averaging 12.9 points and 6.2 rebounds alongside Barry Brown and Kamau Stokes.
He was arguably even better as a junior when he averaged 16.2 points and 4.8 rebounds on a team that reached the Elite Eight.
A series of unfortunate injuries prevented him playing significant minutes in the NCAA Tournament during his final two seasons of college basketball. He has experienced better health in the NBA, and now he is ready to play for his second team.