Bucks sign Jrue Holiday to $135 million extension

FILE – Jrue Holiday #21 of the Milwaukee Bucks handles the ball during a game against the Boston Celtics at Fiserv Forum on March 24, 2021 in Milwaukee, Wisconsin. Stacy Revere/Getty Images/AFP (Photo by Stacy Revere / GETTY IMAGES NORTH AMERICA / Getty Images via AFP)

The NBA’s Milwaukee Bucks have signed guard Jrue Holiday to a contract extension worth a reported $135 million over four years.

The Bucks said Sunday they had agreed to a multi-year extension with Holiday, the guard declining his player option for the 2021-22 season.

ESPN, citing Holiday’s agent, reported the financial terms of the max extension which includes bonuses that could make it worth $160 million.

Holiday arrived in Milwaukee from the New Orleans Pelicans in November as part of a four-team trade.

The two-time All-NBA Defensive Team honoree has appeared in 38 games and is averaging 17.0 points, 5.4 assists, 4.6 rebounds and 1.82 steals in 31.7 minutes per game.

He missed seven games after testing positive for COVID-19 in February.

Since March 20, Holiday has averaged 23.3 points on 58.7% shooting, 7.3 assists, 4.8 rebounds and 1.9 steals in his last eight games.

The announcement comes a day after Holiday scored a season-high 33 points with 11 assists and seven rebounds in the Bucks’ 129-128 victory over the Sacramento Kings.