Forward Dean Wade has agreed to a four-year contract worth $39 million with the Philadelphia 76ers, according to ESPN. Wade, 29, spent the first seven seasons of his career with the Cleveland Cavaliers after going undrafted in 2019.
Wade had been in negotiations with the Cavaliers for a new deal before the start of free agency. However, those talks concluded Tuesday, making him available to sign with other teams. The 76ers were reported to be among several teams interested in Wade last week.
During the 2025-2026 season with the Cavaliers, Wade averaged 5.8 points and 4.2 rebounds per game, with 1.5 assists. He shot 43.9% from the field and 36.2% from 3-point range. He started in 14 of the Cavaliers' 18 playoff games last season, as the team reached the Eastern Conference finals. Despite averaging 4.4 points in 22.6 minutes per game during the playoffs, his defense on opposing stars was considered a valuable asset to the Cavaliers' playoff run. Wade's plus-5.0 net rating during the playoffs was the highest among all Cavaliers players.
His defensive capabilities, combined with his 6-foot-9 height and 6-foot-10 wingspan, made him a consistent presence in Cleveland's frontcourt rotation. Wade's departure makes him the longest-tenured player from his original Cavaliers draft class.