Dean Wade signs four-year $39M deal with Philadelphia 76ers

Free agent forward Dean Wade has agreed to a four-year, $39 million contract with the Philadelphia 76ers, the team's first major move under new president of basketball operations Mike Gansey. The deal was finalized on Tuesday night, the opening day of the NBA's free agency period, after Wade's negotiations with the Cleveland Cavaliers, his team for the past seven seasons, concluded without an agreement.

Wade, 29, established himself as a critical defensive and shooting piece in Cleveland's frontcourt. The 6-foot-9 forward, who went undrafted out of Kansas State in 2019, spent his entire NBA career with the Cavaliers, becoming their longest-tenured player. He averaged 5.8 points and 4.2 rebounds while shooting 36.2 percent from three-point range during the 2025-26 season. His value was most evident on defense, where his size and wingspan made him a versatile perimeter defender capable of guarding opposing stars.

He started 14 of Cleveland's 18 playoff games this past season as the team reached the Eastern Conference finals for the first time in eight years. Wade posted the highest net rating among all Cavaliers players during that postseason run.