Melbourne, Jan. 22 (BNA): World number two Daniel Medvedev beat Dutchman Botic van de Zandschul 6-4, 6-4, 6-2 on Saturday to reach the fourth round of the Australian Open and stay on his way to winning the second title in a row. Hello address.
Reuters said Van de Zandschul, ranked 57th in the world, was the only player to get a boost when the pair met at their previous meeting at the US Open last year, which Medvedev won after claiming his first major title.
Medvedev, the top-ranked player remaining in the men’s draw due to the absence of Serbian Novak Djokovic, made it easy on Saturday, breaking his opponent four times in the match to win under two hours at the Margaret Court Arena.
The 2021 Melbourne Park finalist will meet either Australian Christopher O’Connell or American Maxime Creasy for a place in the quarter-finals.
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