Egyptian international footballer Mohamed Salah has won the 2018 Confederation of African Football Player of the Year award for the second year in a row. The Liverpool forward beat his club teammate, Sadio Mane of Senegal and Arsenal’s Pierre-Emerick Aubameyang, who plays for Gabon, to win the award for the continent’s best player. Also Read - LIV vs LEI Dream11 Team Tips And Predictions, Premier League: Football Prediction Tips For Today’s Liverpool vs Leicester City on November 23 Monday
“I have dreamt of winning this award since I was a child and now I have done so twice in a row,” Salah said at a ceremony here on Tuesday. Also Read - Lionel Messi or Cristiano Ronaldo? Former Liverpool And Man Utd Star Michael Owen Weighs in on Football's Biggest Debate
“I’m proud to win it twice, I must thank my family and my teammates, and I dedicate this award to my country, Egypt”, he said. Also Read - Happy Diwali 2020 Wishes: Virat Kohli Leads Sports Fraternity Wishes on Diwali, Urges Fans 'Not to Burst Crackers'
The 26-year-old scored 44 goals for Liverpool last season as he led the team to the Champions League final. Although he started this campaign slowly following the summer’s World Cup, at which he scored twice for Egypt, he has notched 16 goals in 29 appearances in all competitions for Liverpool so far this season. Along with Salah, Mane, and Aubameyang, Atletico Madrid’s Ghanaian midfielder Thomas Partey was named in the Team of the Year.
That quartet were joined by Manchester United’s Ivorian central defender Eric Bailly, Liverpool and Guinea midfielder Naby Keita, Manchester City’s Algerian winger Riyad Mahrez and Tottenham Hotspur’s Serge Aurier, who represents Côte d’Ivoire.
Meanwhile, Seven-time champions Egypt have won the hosting rights for the 2019 edition of the African Cup of Nations. Egypt beat South Africa by 16 votes to one at the Confederation of African Football’s (CAF) Executive Committee, which was held on Tuesday.
(With IANS Inputs)