Djokovic hnehtu Fonseca a tlawm ta

Kum 19 mi Joao Fonseca chu French Open-a a mumang ram aṭanga kaih harhin a awm ta - quarterfinal-ah champion a beiseina chu a thamral a, amah ang thoa kum naupang ve tak, kum 20 mi Jakub Mensik a ngam lo.

Mensik hian 6-4 6-3 7-6 (7-3) in hnehna a chang a, major tournament-a a vawi khat nana quarterfinal khel pahnih intawngah Mensik a chak zawk - final luh tumin Alexander Zverev a hmachhawn leh ang.

Brazilian tennis star Fonseca hian a quarterfinal kalkawngah hian Novak Djokovic leh Casper Ruud te pawh a paltlang a, hei vang hian a inringtawk hle. Mahse, Mensik hian Alex de Minaur leh Andrey Rublev te a kalkawngah a hneh ve tho a, a inringtawk.

A tawpah chuan Mensik a dingchang a, semifinal a khel dawn - Fonseca leh Mensik inhmachhawn hi tennis khawvela hmabak nei êng pahnih inhmachhawn a nih avangin ngaihventu an tam a, player pahnih hi Jannik Sinner leh Carlos Alcaraz te nen la inel chho ber tura ngaih an ni.

Mensik chuan, "A tawpa hnehna chang zawk ka ni ta chu ka lawm ngei mai. Vawi tam a hnufual zawka ṭan chang ka nei a, mahse, tha ka thlah duh lo a, a tawp thlengin ka bei a ni," a ti a, a tawp thlenga a beih rah a seng thu a sawi.