76ers signing Dean Wade on $39 million contract to boost frontcourt
Dean Wade Signs Four-Year, $39 Million Deal with 76ers
After spending seven years with the Cleveland Cavaliers, free agent forward Dean Wade is making a move to the City of Brotherly Love. According to ESPN, Wade has agreed to a four-year, $39 million deal with the Philadelphia 76ers.
Wade, 29, initially joined the Cavaliers as an undrafted player in 2019, signing a two-way contract with the team. Over the years, he established himself as a critical piece in the frontcourt, known for his defensive abilities and shooting prowess.
Throughout the 2025-26 season, Wade averaged 5.8 points, 4.2 rebounds, and 1.5 assists per game. His shooting percentages were also impressive, with a 43.9% field goal percentage and a 36.2% success rate from beyond the arc.
During the playoffs, Wade started in 14 of the Cavaliers’ 18 games, helping the team reach the Eastern Conference finals for the first time in eight years. Despite averaging 4.4 points in 22.6 minutes per game in the postseason, Wade’s defensive contributions against top players like Scottie Barnes and Cade Cunningham were instrumental in the Cavs’ playoff run.
His stellar performance on the defensive end was reflected in his plus-5.0 net rating, the highest among all Cavaliers players throughout the playoffs.
Philadelphia was among several teams interested in acquiring Wade, with insiders Marc Stein and Jake Fischer predicting the move last week. Wade’s 6-foot-9 height and 6-foot-10 wingspan make him a formidable presence on the court, especially on the defensive end.

Wade’s departure from the Cavaliers marks the end of his tenure as the longest-tenured player on the team. Despite ongoing negotiations for a new deal with Cleveland, Wade ultimately decided to test the free agency market and found a new home with the 76ers.
With his defensive prowess and shooting versatility, Wade is set to bring a valuable skill set to the 76ers as they look to strengthen their roster for the upcoming season.



