Victoria Azarenka of Belarus plays a shot to Barbora Zahlavova Strycova of the Czech Republic during their third round match at the Australian Open tennis championship in Melbourne, Australia, Saturday, Jan. 24, 2015. (AP Photo/Bernat Armangue)
Bernat Armangue
January 24, 2015

MELBOURNE, Australia (AP) Two-time champion Victoria Azarenka moved into the fourth round of the Australian Open on Saturday with a 6-4, 6-4 win over Barbora Zahlavova Strycova of the Czech Republic.

Azarenka, unseeded here because her ranking fell late last year during an injury layoff, beat No. 8-seeded Caroline Wozniacki in the second round after defeating 2013 Australian Open semifinalist Sloane Stephens of the United States in the first.

Azarenka set up match point with a return from the back of the court after first missing a volley from the front.

The 2012 and 2013 Australian Open champion will play last year's runner-up Dominika Cibulkova in the fourth round on Monday.

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