The second-seeded Indian got the better of Vietnam's Tien Minh Nguyen in straight games.
Indian shuttler Sourabh Verma brushed off a stern challenge from Tien Minh Nguyen to book his place in the semi-finals of the Vietnam Open on Friday.
Beating the home hero 21-13, 21-18 in straight games, Verma - who won the Hyderabad Open last month - will lock horns with Minoru Koga of Japan in the last four of the competition.
Following the elimination of Shubhankar Dey, Siddharth Pratap Singh and Siril Verma earlier in the tournament, Sourabh Verma continued his progress as India’s lone hope in absolute style, having defeated Japan’s Yu Igarashi 25-23, 24-22 in a hard-fought encounter on Thursday.
On Friday, the second-seeded Indian, who received a bye into the second round of the Open, got off to a nervy start.
Despite having an inferior head to head record against his Vietnamese opponent, Verma surged into a 12-6 advantage in the first game.
From there, he allowed his opponent only seven more points before grabbing the game 21-13.
The second game, however, saw the Vietnamese breathing right down Verma’s neck as Nguyen never really allowed his Indian counterpart to enjoy a comfortable lead.
As the game approached its second break, the two shuttlers were tied tight at 10-10.
A quick-fire barrage of four consecutive points though suddenly put Verma in the driving seat.
Nguyen, on his part, reacted in determined fashion to close down the gap to just a point. With the game now hanging at 19-18, Verma grabbed two consecutive points to book his place in the semi-final.