New Delhi: Thailand shuttler Supanida Katethong defeated Indian ace shuttler PV Sindhu 2-1 in the women’s semifinals of the India Open 2022 at the KD Jadhav Stadium, New Delhi, on Saturday, January 15, 2022.
Sindhu opened the set with a point immediately equalised by Supanida, who showed a brilliant start and took a 6 point lead. However, Sindhu took advantage of Supanida’s back-to-back errors and tried equalising scores.
However, Sindhu’s unforced errors left her with a 6-point disadvantage. Sindhu dominated the match again and tried levelling scores. She made an angular court smash as Supanida fumbled. Despite expectations, Supanida bounced back and defeated Sindhu in the first set by 21-14.
The next set began with the Thai in charge of the match as she gained a quick point. A drop shot by Supanida gave Sindhu an early advantage to equalise scores. Sindhu took over the domination and was in the lead with a 12-10 in no time. Supanida’s unforced errors and faults did not stop her from trailing a point behind.
Sindhu outsmarted Supanida into gaining a 7-point lead and increased the distance. The set ended with Sindhu winning 21-13.
The final set began with both players at an equal footing. The decider match had Sindhu in the lead as Supanida caught up and equalised scores at 4-4. Supanida made a down-the-line shot and a clear 4-point lead as the Thai picked a baseline shutter as Sindhu faltered. Sindhu gave another 4-point lead to the Thai when she made another service error.
Meanwhile, Supanida’s forehand smashes, swift movements, court coverage, and aces gave her another 9-point advantage before she won the semifinals against the Olympian 21-10. The intriguing match lasted 56 minutes.
Supanida Katethong will face her Thai compatriot Busanan Ongbamrungphan in the India Open 2022 finals at the KD Jadhav Stadium, New Delhi, on Sunday, January 16, 2022.
(Avya Mathur)
