In addition to signing Jordan Bell, the Cavs also converted Dean Wade’s two-way contract to a multi-year deal on Monday. The move, first reported by ESPN’s Adrian Wojnarowski, will pay Wade a sum for the rest of the season and is non-guaranteed beyond this year, per league sources.
The structure of Wade’s deal, per sources, is similar to that of Bell and Alfonzo McKinnie’s deals. Basically, the Cavs pay him now, lock him in for camp and can see if they want to keep him going forward. Wade, who went undrafted in the 2019 NBA Draft partly due to injuries, had an interesting season with the Canton Charge. He shot just under 40% on three-pointers while averaging 14.2 points and 7.5 rebounds per game.
With the Cavs, Wade would provide some frontcourt depth as a stretch four who is decently mobile on defense. Consistent minutes for him could be tough as long as Kevin Love is a Cavalier, but he’s an interesting prospect who won’t cost the team all that much to get an extended look at. Per cleveland.com, the Wizards also had interest in Wade. Had the Cavs not converted his deal, he would have been a free agent at the end of the restarted season.
Wade’s deal and Bell’s also aren’t so big that they push the Cavs above the luxury tax threshold this year, which makes sense for a whole bunch of reasons.