“The light will play against the driver, Ozh-allyasim will meet with the ruble”, – WRITE: Sport.ua
The first sown Carlos Alkaras gave way to Girzhi. The Czech for the first time in his career overcame the top 3 rating of the ATR and for the second time in the season will play in the semifinals.
The next Czech rival will be British Jack Draper. He overcame the abuser Novak Dzhokovich Matteo Berretin. For the first time in October, Draper came to the semifinals at the ATR level. Then he took the trophy in Vienna.
In another semi-final confrontation, Felix will come together and Andrey Rublev. The Canadian took the first set from Daniel Medvedev, after which “Neutral” starred.
By the way, for Felix is the second refusal of a rival for the tournament. Hamad Medzedovich starred from the 2nd round. The Serb at the start of the competition on one leg passed Stephenos Cycations, but failed to go to his next match.
As for Rublev, which, unfortunately, still plays in Qatar and has not yet shown his psychic, beat the second racket of Alex de Minaura tournament.
The semi -finals will still take place on Friday, February 21.
ATR 500 Doha. 1/8 finals
The upper part of the grid
Carlos Alaras [1] – 
Nardy bow – 6: 1, 4: 6, 6: 3
Fabian Mrozhan – 
Rusty League – 4: 6, 2: 6
Matteo Berrettini – 
Tallon Grickspor – 7: 6 (7: 4), 6: 7 (6: 8), 6: 4
Christopher O’Connel – 
Jack Draper [8] – 2: 6, 1: 6
The lower part of the grid
Hamad Medzedovich – 
Felix is an all-aliasim-wo. Over-ali
Zizu Bergs – Danil Medvedev [4] – 2: 6, 1: 6
Andrey Rublev [5] – 
Nuno Borzhe – 6: 3, 6: 4
Botic van de Zandshulp – Alex de Minaur [2]
ATR 500 Doha. 1/4 finals
The upper part of the grid
Carlos Alaras [1] – Rusty League – 3: 6, 6: 3, 4: 6
Matteo Berrettini – Jack Draper [8] – 6: 4, 4: 6, 3: 6
The lower part of the grid
Felix Ozh-Alyasim-Danil Medvedev [4] – 6: 3, Ret., Medvedev
Andrey Rublev [5] – Alex de Minaur [2] – 6: 1, 3: 6, 7: 6 (10: 8)
ATR 500 Doha. Pairs 1/2 finals
Rust Legiechka – Jack Draper [8]
Felix Ozh-Alyasim-Andrey Rublev [5]