© Kyle Terada | 2015 Dec 25
The Golden State Warriors have signed center Andrew Bogut to a contract, the team announced today. Per team policy, terms of the agreement were not released.
Bogut, 34, recently concluded the 2018-19 season with the Sydney Kings of the National Basketball League (NBL) in Australia, where he earned the league’s Most Valuable Player, Defensive Player of the Year and All-NBL First Team honors after averaging 11.4 points, a league-high 11.6 rebounds, 3.4 assists and 2.7 blocks in 29.7 minutes over 30 games.
A 13-year NBA veteran, Bogut joins the Warriors for a second stint after previously spending four seasons in Golden State (2012-16), where he was a key member of the 2014-15 championship team and helped the Warriors set an NBA record for single-season wins (73) in 2015-16. In four seasons with the Warriors, Bogut averaged 6.3 points, 8.3 rebounds, 2.2 assists and 1.71 blocks in 23.7 minutes over 236 games (230 starts). He ranks second on Golden State’s all-time field goal percentage list (.581) and was named to the NBA All-Defensive Second Team in 2014-15.
Bogut owns career averages of 9.8 points, 8.7 rebounds, 2.2 assists and 1.56 blocks in 28.4 minutes over 694 games (656 starts) with the Milwaukee Bucks, Warriors, Dallas Mavericks, Cleveland Cavaliers and Los Angeles Lakers. He last appeared in an NBA game with the Lakers last season on Jan. 1, 2018.
Originally selected by Milwaukee with the first overall pick in the 2005 NBA Draft, Bogut was named to the NBA All-Rookie First Team in 2006 and was an All-NBA Third Team selection in 2009-10. In 58 career postseason appearances (57 starts) with Milwaukee and Golden State, Bogut owns averages of 5.5 points, 7.6 rebounds, 1.8 assists and 1.50 blocks in 22.5 minutes. He ranks second on Golden State’s career postseason field goal percentage list (.589) and third on the franchise’s career postseason blocks per game list.
