Radu Albot qualified for the semi-finals of the ATP Challenger “Moldova Open” tournament after defeating Italy’s Franco Agamenone in the quarter-finals, IPN reports.
The match ended 7-5, 7-6 in favour of the Moldovan tennis player. Albot won the first set after a balanced duel, while in the second he prevailed in the tie-break. According to match statistics, the Moldovan scored 85 points, compared with his opponent’s 76, and converted five of the 16 break points he created. Albot also won 41 return points, compared with the Italian’s 26.
In the semi-finals of the competition, which is taking place in Chisinau, Radu Albot will face Romania’s Cezar Cretu on Saturday. Earlier, the Moldovan tennis player defeated France’s Maxime Janvier and Kazakhstan’s Timofey Skatov.