Dean Wade Joins the 76ers

Dean Wade is officially heading to Philadelphia.

The forward, now a free agent, has signed a four-year contract worth $39 million with the 76ers, as reported recently.

At 29, Wade is parting ways with the Cavaliers, where he made a name for himself as a reliable defender and shooter in the frontcourt. Surprisingly, he went undrafted initially and played his first seven seasons in Cleveland after joining the team on a two-way contract back in 2019.

Interestingly, Wade was the longest-tenured player on the Cavaliers’ roster.

Negotiations for a new contract were ongoing between Wade and Cleveland until Tuesday, just before free agency opened. However, by the afternoon, it became clear he would enter the free-agent market, prompting the 76ers to scout him right away.

NBA insiders, including Marc Stein and Jake Fischer, had mentioned last week that Philadelphia was among several teams interested in signing Wade.

Known for his defensive skills, Wade carved out a significant role in Cleveland’s lineup. At 6 feet 9 inches tall, with a wingspan to match, he posed a challenge for opposing players.

During the 2025-26 season, Wade averaged 5.8 points, 4.2 rebounds, and 1.5 assists per game, with shooting percentages of 43.9% overall and 36.2% from beyond the arc.

He played a crucial role in the playoffs last season, starting in 14 of the Cavaliers’ 18 games and helping the team reach the Eastern Conference finals for the first time in eight years, though they eventually fell to the Knicks.

Throughout the playoffs, Wade averaged 4.4 points in about 22 minutes per game, but his defensive contributions against top players like Scottie Barnes and Cade Cunningham were invaluable.

Notably, his plus-5.0 net rating was the highest among all Cavaliers during the playoffs.