Jrue Holiday has reportedly agreed to a four-year maximum contract extension with the Bucks worth up to $160 million, according to The Athletic's Shams Charania.
The 30-year-old guard joined Milwaukee from New Orleans in November in a four-team trade that saw the Bucks part with Eric Bledsoe, George Hill, three future first-round picks and two future draft pick swaps.
Holiday is currently averaging 17 points and 5.4 assists per game this season and is second in the NBA with 1.8 steals per game. The former UCLA standout was an All-Star in 2013 and won NBA Teammate of the Year last season.
With the extension, the Bucks lock up a member of their core who had a player option for $27 million in 2021-22, which could have made him a free agent in the summer if he had opted out.
The Bucks now have Holiday, Giannis Antetokounmpo and Khris Middleton on the books at upwards of $100 million for at least the next two seasons as they search for their first NBA Finals appearance since 1974.
Last night, Holiday scored 33 points to lead Milwaukee, which was without Antetokounmpo due to a sore knee, to a 129–128 victory over the Kings.
The extension also wraps up a busy week for the Bucks, who are third in the Eastern Conference, after they signed free agent Jeff Teague Thursday. The veteran guard was recently waived by the Magic after the Celtics traded him to Orlando at the trade deadline.