News & Events in Belarus

MINSK, 9 January (BelTA) – Belarus number one Victoria Azarenka won the Brisbane International in Australia after beating Germany’s Angelique Kerber, ranked 10^th in the world, 6-3, 6-1, BelTA informs.

Azarenka went strong all week, losing just 13 games in her four matches leading up to the final. She maintained that form against the No.4-seeded Angelique Kerber on Saturday night. Azarenka finished with a +14 differential of winners to unforced errors, 23 to 9, while Kerber was a slightly negative -2 on winners to unforced errors, 25 to 27. The match lasted for 74 minutes.

The victory brought Victoria Azarenka $193,000 and 479 points, which would take her back to the top 20. It is Azarenka’s second triumph in Brisbane since 2009 when she beat Marion Bartoli of France 6-3, 6-1. In 2014 the Belarusian lost the title to Serena Williams of the United States 4-6, 5-7. The 26-year-old Belarusian has won 18 WTA titles in her career so far. Last time she celebrated a victory was in Cincinnati, the United States in August 2013 when she beat Serena Williams.

On her way to the Brisbane final Victoria Azarenka saw off Russia’s Elena Vesnina (World No.115) 6-2, 6-0. With Simona Halep withdrawing, Azarenka progressed to the third round where she defeated Belgium’s Ysaline Bonaventure (170) 6-3, 6-2. In the quarterfinal the Belarusian was too strong for World No.15 Roberta Vinci of Italy 6-1, 6-2. In the semifinal Azarenka beat Samantha Crawford of the United States (142) 6-0, 6-3.

Now, the world’s leading tennis players are gearing up for the Australia Open which is due in Melbourne on 18-31 January. Victoria Azarenka won the Australian Open twice: in 2012 and 2013. Last year the Australian Open title went to Serena Williams. China’s Na Li celebrated the victory in 2014.