Indian shuttler Sourabh Verma stormed into the men’s singles semifinals of the Vietnam Open BWF Tour Super 100 after beating local hero Tien Minh Nguyen in straight games in Ho Chi Minh City on Friday.
Second seed Sourabh, who had won the Hyderabad Open last month, defeated Tien Minh 21-13, 21-18 in a 43-minute clash.
The reigning national champion will face Japan’s Minoru Koga who upset Thailand’s sixth seed Tanongsak Saensomboonsuk 13-21, 24-22, 21-16 in a 72-minute marathon encounter.
Sourabh jumped to a 4-1 lead early on and then entered the break with a 11-6 advantage. The Indian kept the proceedings under control after the interval to eventually pocket the opening game comfortably.
The second game was more competitive as Sourabh and Tien Minh exchanged the lead frequently before the Vietnamese shuttler managed a slender 11-10 lead at the breather.
However, Sourabh soon turned the tables, moving to 17-12. Tien Minh managed to narrow the deficit to 18-19 but the Indian reeled off the required two points to shut the door on the 35-year-old Vietnamese.
The 26-year-old Indian had lost in the round of 16 at the Chinese Taipei Open earlier this month. In the third meeting between the two, the Indian national champion emerged triumphant for the first time. Tien Minh, former world No 5, had defeated Sourabh at Singapore Open earlier this year.
(With PTI inputs)
