Jordan Ayew worked very hard in the match against Southampton


Ghana striker Jordan Ayew was named the best player for English side Aston Villa on Saturday even though they were not able to grab victory at Southampton.

The victors looked like they were coasting to their first victory in 14 matches but Oriol Romeu’s late equaliser earned Southampton a draw.

This means Aston Villa side who have now gone 14 Premier League games without a win but still the Ghanaian attacker Ayew was hailed once again as the best player for the side.

The Ghanaian rated 7.5 out of the 10 marks after the game, making him the best player for the side.

“No goal this time but his work rate was outstanding once again, while he looked a constant threat with the ball. Should be the first name on the team sheet at the moment,” the statement on Ayew’s performance hailed the Ghanaian.

The Midlands club are yet to win in four games under manager Remi Garde and they would have been on the end of a heavy defeat if Southampton had been more clinical in front of goal.


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