Dean Wade Is Leaving Cleveland For Philadelphia On A Four-Year $39 Million Deal

Dean Wade agreed to a four-year, $39 million deal with the Philadelphia 76ers on Monday night, leaving the Cleveland Cavaliers after five seasons to join a 76ers team that is rebuilding around Joel Embiid and a new direction under first-year head coach Ime Udoka.

The deal averages $9.75 million per year, a significant raise for a player who had been earning roughly $8 million annually in Cleveland and who has established himself as one of the more reliable stretch forwards in the league.

Wade, 27, from Millbrook, Alabama, went undrafted out of Kansas State in 2019 and spent his entire NBA career in Cleveland after signing as a free agent. Over five seasons with the Cavaliers he developed from a two-way player fighting for roster spots into a dependable rotation piece averaging 8.4 points, 4.2 rebounds and 1.1 steals per game last season while shooting 38.9 percent from three on four-plus attempts.

His value is straightforward, he is a 6-foot-9 forward who can shoot the ball from the perimeter, defend multiple positions and not turn it over.

In Cleveland's system alongside Donovan Mitchell and Darius Garland, those qualities were sometimes overlooked.

In Philadelphia, where Udoka is building a new culture and needs veterans who do the right things without being asked twice, they land differently.

The 76ers have had an active start to free agency. Wade joins a Philadelphia team that still has Embiid under a max contract and is adding around him rather than blowing things up entirely.

A four-year deal signals the organization sees Wade as part of whatever comes next, not just a stopgap.