Forward Dean Wade continued his steady rise with the Cavaliers as the team officially announced his three-year, $18.5 million contract extension that was agreed to on Saturday.
The Cavs had previously picked up Wade's fourth-year option on June 22. A league source said $16.5 million of that is fully guaranteed as the extension keeps Wade in Cleveland through the 2025-26 season.
A two-time All-Big 12 player during his four years at Kansas State, Wade went undrafted, joining the Cavs in 2019-20 on a two-way deal. He saw action in just 12 games as a rookie, but impressed with his 3-point shooting, rebounding and toughness.
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Last season, Wade played in 51 games and started 28 before undergoing surgery on March 28 to repair a torn meniscus in his right knee. Wade averaged 5.3 points, 2.9 rebounds and 1.0 assists, pushing that to 7.8 points, 3.8 rebounds and 1.5 assists as a starter.
He scored in double figures in 11 of those 28 and shot 42.9% from long range in his final 12 games as a starter from Jan. 24 through March 12.
Wade missed the final 15 games, sidelined along with All-Star center Jarrett Allen down the stretch, as the Cavs (44-38) were eliminated with losses to the Brooklyn Nets and Atlanta Hawks in the play-in tournament.
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Boosted by the blockbuster Sept. 1 trade for four-time All-Star guard Donovan Mitchell, the Cavs open training camp with Monday's media day at Rocket Mortgage Fieldhouse. Wade is among those vying for the small forward spot, along with Isaac Okoro, Caris LeVert and Cedi Osman.
This article originally appeared on Akron Beacon Journal: Cavs forward Dean Wade agrees to three-year contract extension