MELBOURNE, 24 January 2020 – Marin Cilic has reached the last 16 at the Australian Open after knocking out world no. 9 Roberto Bautista Agut from Spain on Friday in Melbourne.
The 31-year-old Croatian hit 24 aces to beat Bautista Agut in five sets 6-7, 6-4, 6-0, 5-7, 6-3 in 4 hours and 10 minutes on Melbourne Arena this afternoon.
After losing the opening set in a tie-breaker, Cilic would get the decisive break in the 9th game of the second set when Bautista Agut made a forehand error. He would hold his serve to take the set 6-4 and level at 1-1.
It was one-way traffic in the third set as Bautista Agut struggled with the power of Cilic. The Croatian would win 100% of points when his first serve went in and cruised to take the set to love.
Cilic broke early in the 4th set and raced to a 3-1 but the Spaniard returned the break to in the 6th game. Cilic had the chance to break again at 5-5 but it would be Bautista Agut who would break at 6-5 and he would force the match into a fifth set.
A forehand winner would give Cilic the break in the opening game of the fifth set. Cilic had a break point at 4-2 but could not convert. He would get the break again though in Bautista Agut’s next service game to take the set and match.
The win was a satisfying one for the Croatian as it was Bautista Agut who knocked him out of the Australian Open in 2019 in a five-set thriller. Cilic’s best performance at the Australian Open came in 2018 when he was runner-up. He then became the first Croatian to ever reach the singles final at Melbourne Park.
A 6-0 blitz in the third!
— #AusOpen (@AustralianOpen) January 24, 2020
Cilic will player either Stefanos Tsitsipas from Greece or Canada’s Milos Raonic in the fourth round.