CLEVELAND, Ohio -- If the Cavs are going to extend their season-high winning streak to 10 games on Monday night, they will need to do it without a key piece of their usual rotation.
Do-it-all-forward Dean Wade, who received the team’s coveted Junkyard Dog award following this weekend’s 119-95 slaughter of the Toronto Raptors, will not play against the Philadelphia 76ers because of an illness.