The Milwaukee Bucks and point guard Jrue Holiday have agreed to a four-year maximum contract extension worth up to $160 million on Sunday, according to multiple reports.
Holiday’s agent, Jason Glushon, confirmed the terms of the deal, per the report.
Holiday, 30, is averaging 17.0 points, 5.4 assists, 4.6 rebounds and a league-leading 1.8 steals in 38 games (35 starts) in his first season with the Bucks.
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Milwaukee acquired the two-time NBA All-Defensive Team selection from the New Orleans Pelicans in a four-team trade in November. The Bucks traded away Eric Bledsoe, George Hill, three future first-round picks and two future draft pick swaps in the deal that also involved the Denver Nuggets and Oklahoma City Thunder.
A first-round selection (17th overall) by Philadelphia in 2009 and an All-Star in 2012-13, Holiday has averaged 15.9 points, 6.3 assists, 3.9 rebounds and 1.5 steals in 751 career games (675 starts) with the 76ers (2009-13), Pelicans (2013-20) and Bucks.
He was voted the NBA’s Twyman-Stokes Teammate of the Year in 2019-20.
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The Bucks entered Sunday’s action in third place in the Eastern Conference with a 32-17 record, two games back of co-leaders Philadelphia and the Brooklyn Nets.
Milwaukee’s top three players are now locked into long-term deals. The Bucks gave two-time MVP forward Giannis Antetokounmpo a five-year, $228.2 million supermax extension in December, and re-signed two-time All-Star forward Khris Middleton to a five-year, $177.5 million deal in July 2019.
–Field Level Media