Dominican boxer Alexy de la Cruz advanced to the quarterfinals of the World Boxing Championship in Belgrade, winning his third fight of the tourn
Dominican boxer Alexy de la Cruz advanced to the quarterfinals of the World Boxing Championship in Belgrade, winning his third fight of the tournament.
De la Cruz, of the lightweight (60 kilos) gave a 5-0 account of Abdurahmon Yoqubov, of Tajikistan, in the third round of the tournament that takes place in the Serbian capital.
Now, the Caribbean fighter will face the Finn Arslan Khataev, in a fight scheduled to take place this Tuesday. The winner of that fight will advance to the semifinal.
De la Cruz, who reached the second round at the last Tokyo Olympics, opened his participation in the tournament with victory in Kenyan Nicholas Okoth, then defeated German Murat Yildirim before his victory on Sunday.
De la Cruz, 25, is a bronze medalist at the Lima 2019 Pan American Games. In the last Tokyo Olympics he lost in the round of 16 to the Russian Albert Batyrgaziev, who eventually won the gold medal.
The World Championship was scheduled to be held in the Indian city of New Delhi, but in April 2020 the International Boxing Association (AIBA) withdrew the venue and granted it to Belgrade.
In a statement, AIBA said it had terminated New Delhi’s contract after it failed to pay an agreed host fee, meaning India will have to pay a cancellation penalty of $ 500,000 (€ 461,366).
The Boxing Federation of India (BFI) claimed that it was unable to pay the corresponding fee because AIBA’s bank account in Lausanne was “frozen.”