Lucas Pouille's fine Australian Open run continued as he reached the semifinals with four-set win over Milos Raonic on Wednesday.
Pouille moved into the last-four of a grand slam for the first time with a 7-6 (7-4) 6-3 6-7 (2-7) 6-4 victory after three hours and two minutes on Rod Laver Arena.
The French 28th seed had lost in the first round in five straight years in Melbourne but he is embarking on the best run of his career this time around.
Pouille – a quarter-finalist at Wimbledon and the US Open in 2016 – will face Novak Djokovic in the last-four.
The loss, Raonic's first in four meetings with Pouille, was the Canadian 16th seed's fourth straight at this stage of a major.
Raonic started brightly and grabbed an early break on his way to a 3-0 lead in the first set but Pouille hung in there and his opponent was unable to serve out, broken to 15 at 5-3.
A double fault from Raonic gave Pouille control of the tie-break and he made the most of the opportunity, a brilliant forehand pass down the line seeing him take the first set.
Pouille was well on top and Raonic had few answers, the Frenchman breaking again for a 3-1 lead in the second after a wonderful lob.
Making few errors, Pouille lost just four points on serve in the set as he took it 6-3.
He continued to dominate and created two break points in the seventh game of the third, but some big serving got Raonic out of trouble.
Perhaps with a first grand slam semi-final appearance on his mind, Pouille played some nervy points late in the set as the duo went to a second tie-break.
Raonic made the perfect start – highlighted by a forehand return winner cross-court for 3-0 – as he won the first six points on his way to pulling himself back into the contest.
Things remained tight until the 10th game of the fourth, Pouille claiming victory when Raonic – having saved two match points – sent a backhand well wide on the third.
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