Chelsea’s Eden Hazard has pipped Manchester City midfielder Kevin De Bruyne to win the prestigious Best Belgian player in Foreign League award. Hazard saw off competition from the talismanic midfielder alongside Dries Mertens who plays for Serie A giants Napoli.
He beat his former teammate De Bruyne by 11 after he landed the accolade by garnering 326 votes. Mertens trailed Chelsea forward by 56 votes as Manchester United striker Romelu Lukaku came at fourth position.
Eden Hazard became the favourite after helping Chelsea to lift the English Premier League crown last year. The Belgian forward played a major role in the campaign as the champions strolled to the historic win by a record 93 points.
He formed formidable partnership with Pedro, Diego Costa and Fabregas making the premier league matches easier to win. The Blues established a 13 game winning streak of which hazard was part of as they matched to the crown.
Hazard scored 14 goals last season before being injured while on the international due with Belgium. While speaking about being voted for the award, he emphasized this will not be his last time to scoop an award from Belgium football stakeholders.
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“My first prize in Belgium and certainly not my last one.” Said Hazard.
He however appeared selfless by hinting that he would have voted for Napoli’s Dries Mertens for staging impressive performance in Italy.
“What a good year he had. With a lot of goals too. I go for him, because he also played in a position that was not actually his. He was not an attacker.” He added.
Kevin De Bruyne has won this award twice. He may be looking to win it a third time after playing a key role in Man City’s title bid.
