Oscar De La Hoya has compared an unbeaten world champion to the legendary Floyd Mayweather.
‘Money’ Mayweather will go down as being one of the greatest fighters of all time, proving his outstanding credentials throughout the course of his glittering professional career.
He reigned as a world champion in five weight classes during his tenure, which began back in 1996 just a few months after he captured a bronze medal at the Olympic Games in Atlanta, Georgia.
Mayweather picked up a number of career-defining victories along the way, defeating the likes of Miguel Cotto, Canelo Alvarez and Manny Pacquiao amongst many others.
Another memorable win for ‘Money’ came against De La Hoya, who he defeated via split decision when they locked horns for the WBC light-middleweight title back in May of 2007.
Speaking to Mill City Boxing, De La Hoya previewed the upcoming lightweight clash between Shakur Stevenson and William Zepeda, where he claims Stevenson is ‘exactly like’ former rival Mayweather.
“So here’s my expert opinion on Shakur Stevenson versus William Zepeda, Saturday night, Ring Magazine event, it’s going to be incredible, New York. Alright, Shakur Stevenson, what type of fighter is he? Exactly like Floyd Mayweather.
“He knows how to box, beautiful footwork, stands up straight, shoulder roll all that good stuff right? Has a beautiful jab, pop shots you with a right hand. Well guess what? Floyd Mayweather lost against Jose Luis Castillo, which reminds me of William Zepeda, but throws more punches.”
Stevenson is set to make the latest defence of his WBC lightweight title against William Zepeda at the Louis Armstrong Stadium in New York City this Saturday night, as he looks to solidify his status as being the best lightweight on the planet.
