Frenchmen Pierre-Hugues Herbert/Nicolas Mahut reached their second Australian Open semi-final on Wednesday, ending Bob Bryan/Mike Bryan’s quest for a 17th Grand Slam title as a team.
Herbert/Mahut fell down a break in the second set, and the Bryans served to level the quarter-final. But Herbert/Mahut broke back in the ninth game and pulled away in the tie-break to advance 6-4, 7-6(3).
The Bryans were playing their first Slam together since last year’s Australian Open, since Bob missed the final six and half months of the 2018 season because of a right hip injury that required surgery.
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Herbert/Mahut made the 2015 final but lost to Italians Simone Bolelli/Fabio Fognini. The Frenchmen will next meet Americans Ryan Harrison/Sam Querrey, who upset seventh seeds Lukasz Kubot/Horacio Zeballos 3-6, 7-6(5), 6-4.
On the top half of the draw, 2017 Australian Open champions Henri Kontinen/John Peers won 88 per cent of their first-serve points (29/33) and knocked out third seeds Jamie Murray/Bruno Soares 6-3, 6-4.
The 12th seeds will play Leonardo Mayer/Joao Sousa for a chance to return to the final. Mayer/Sousa beat sixth seeds Raven Klaasen/Michael Venus on Tuesday.
Source: ATP World Tour
Herbert/Mahut End Bryans' Run In Melbourne
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