Dwight Howard Reportedly Expected to Re-Sign with Los Angeles Lakers
According to Sam Amick of USA Today, two sources close to the situation fully expect Howard to sign a long-term extension with Los Angeles. While his final decision won’t be made until the summer, as of now, it appears the All-Star center will remain in the Purple and Gold for years to come.
For Lakers supporters, this information couldn’t come at a better time.
After the devastating news of Kobe Bryant’s Achilles' injury—which will keep him sidelined for the next six to nine months—the Lakers front office has to feel good knowing one of its superstars will remain on the roster for the foreseeable future.
Per Amick, Los Angeles can put a five-year, $117.9 million extension on the table this coming summer, while other teams can only offer a four-year, $87.8 million deal at the most. The extra season represents an extra $30 million of financial security for Howard, who turned 27 years old in December.
In addition, the Lakers will essentially be Howard’s team with Bryant sidelined, and there’s a good chance things remain that way if and when the Black Mamba finally returns.
Howard has already proven that he’s up for the challenge, leading the Lakers to a crucial win over San Antonio on Sunday by scoring 26 points on 9-of-15 shooting to go with 17 rebounds and three blocked shots.
Of course, even if Howard plays well and leads Los Angeles into the postseason, it won’t stop a number of organizations from clearing cap room and attempting to woo the star center in free agency.
Where will Howard sign?
While those teams will have their work cut out for them bidding against the Lakers, much could change between now and June, especially if Los Angeles misses the playoffs or suffers a disappointing first-round sweep.
Considering the "Dwightmare" that took place during the 2011-12 season, Howard's free-agent decision cannot be determined with certainty at this time. With that said, don’t be surprised if he ultimately commits to the Lakers for the next half decade.
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