Dean Wade is heading to the City of Brotherly Love.

Per ESPN’s Shams Charania, the free agent forward has agreed to a four-year, $39 million deal with the 76ers.

Wade is leaving the Cavaliers, where he established himself as a critical defending and shooting piece in the frontcourt. Wade spent the first seven years of his career in Cleveland after going undrafted and signing a two-way contract with the team in 2019. Wade had been the longest tenured Cavalier.

Until Tuesday, the Cavs and Wade had been having ongoing negotiations trying to work out a new deal before the start of free agency. By late Tuesday afternoon, it became clear that Wade would hit the free agency market, and now Philly scoops him up on the first day.

NBA insiders Marc Stein and Jake Fischer reported that Philadelphia was expected to be one of several teams interested in acquiring Wade last week.

Wade’s defense made him an important piece of Cleveland’s rotation and allowed him to retain a consistent role there. His 6’9” height and 6’10” wingspan made it difficult for opposing guards and forwards to work around him.

Wade averaged 5.8 points and 4.2 rebounds alongside 1.5 assists per game during the 2025-26 season, shooting 43.9 percent from the field and 36.2 percent from three-point territory.

Wade started in 14 of Cleveland’s 18 playoff games last season, where they reached their first Eastern Conference Finals in eight years before being swept by the Knicks. Wade only put up 4.4 points in 22.6 minutes per game in the playoffs, but his defense on stars like Scottie Barnes and Cade Cunningham made him a valuable piece of the Cavs’ run.

His plus-5.0 net rating was the highest among all Cavs players throughout the playoffs.