Vietnam Open 2019: Shuttler Sourabh Verma Reaches Semis After Beating Tien Minh Nguyen in Straight Games

Ho Chi Minh City, September 13: India shuttler Sourabh Verma got the better of home favourite Tien Minh Nguyen to enter the men's singles semi-final at the Vietnam Open BWF Tour Super 100 tournament in Ho Chi Minh City.

Sourabh beat Tien Minh 21-13 21-18 in a 43-minute clash on Friday to set up a semifinal clash against Japan's Minoru Koga, who beat Thailand's sixth seed Tanongsak Saensomboonsuk 21-13, 21-18 later in the day.

Verma, the reigning national champion, got a bye in the first round and beat Japan's Kodai Naraoka 22-20, 22-20 in the second. He became the lone Indian left in the tournament following the ouster of Siril Verma and Shubhankar Dey earlier on Thursday.

Sourabh raced to a 4-1 lead early on and kept his foot on the pedal to reach the interval leading 11-6. He continued in the same vein to record a straight forward win in the first game. The second game was a more level affair and it was Tien Minh who held a slended lead of 11-10 at the interval.

Sourabh then stormed to a 17-12 lead before the Vietnamese narrowed the lead down to 19-18. Sourabh kept his cool and did not allow his opponent to score any more points after that.

(The above story first appeared on LatestLY on Sep 13, 2019 08:46 PM IST. For more news and updates on politics, world, sports, entertainment and lifestyle, log on to our website latestly.com).