Britain’s Amir Khan will defend his world light-welterweight titles against US mandatory challenger Lamont Peterson on December 10 in Washington, it was confirmed Tuesday. Khan, 26-1 with 18 knockouts, has won eight fights in a row, most recently last July in Las Vegas when he knocked out Zab Judah in the fifth round to unify the World Boxing Association and International Boxing Federation titles. “I am looking forward to returning to the United States to face a worthy challenger like Lamont Peterson,” Khan said in a statement. “I always said I wanted to be active and a true world champion and I will prove it on December 10. “I know Lamont is a strong contender but it doesn’t matter that I’m fighting him in his back yard, I’m coming in as champion and will leave as champion.” Peterson, 29-1-1 with 15 knockouts, fought Victor Ortiz to a 10-round draw last December on the undercard of Khan’s unanimous decision victory over Marcos Maidana. The fight is expected to be the last for the Englishman in the division before he moves up next year to welterweight, where possible opponents could include unbeaten US star Floyd Mayweather and Filipino icon Manny Pacquiao. Peterson’s only loss came in his only prior world title fight, when he dropped a unanimous 12-round decision in December of 2009 to Tim Bradley.
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