CLEVELAND - Dwyane Wade is headed to Cleveland. The 12-time All-Star and 3-time NBA champion will sign with the Cleveland Cavaliers when he clears waivers on Wednesday, ESPN's Adrian Wojnarowski reports.
On Monday, Cavs forward LeBron James admitted he would attempt to recruit Wade to Cleveland after his friend and former teammate reached a buyout agreement with the Chicago Bulls over the weekend. James spent four seasons playing alongside Wade with the Miami Heat, with the superstar duo reaching four straight NBA Finals and winning two championships from 2011-14.
One of the most decorated shooting guards in NBA history, the 6-foot-4 Wade has averaged 23.3 points, 5.7 assists and 4.8 rebounds over the course of his 14-year career. After spending the first 13 years of his career in Miami, the 2006 NBA Finals MVP signed with the Bulls in 2016 and averaged 18.3 points in 60 games with Chicago.
"I would love to have D-Wade a part of this team," James said at Cavs media day on Monday. "I think he brings another championship DNA, championship pedigree. He brings another playmaker to the team who can get guys involved, and make plays, and also just has a great basketball mind."
In adding Wade, Cleveland bolsters a backcourt in need of offensive firepower following the trade that sent Kyrie Irving to the Boston Celtics in exchange for guard Isaiah Thomas, who is currently recovering from a hip injury.
Asked how adding Wade could potentially affect his own role, the Cavs' current starting shooting guard, J.R. Smith, said "I don't care, as long as we win."
According to Yahoo Sports' Shams Charania, Wade will sign a one-year, $2.3 million deal with Cleveland.
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