Saturday Oct 27, 2018. 10:00
Moroccan star Hakim Ziyech is set to make his 100th official appearance for AFC Ajax when they take on Feyenoord in a crunch Eredivisie match on Sunday.
‘De Klassieker’ (The Classic) between Ajax and Feyenoord is one of the greatest rivalries in Dutch football and Ziyech will be honoured to bring up his century in a match of such prestige, especially in front of his home fans at the Johan Cruyff Stadium.
The Dutch-born 25-year-old joined Ajax from FC Twente in August 2016 and has become one of De Godenzonen’s key players.
His 99 competitive appearances have seen him rack up 26 goals and 42 assists – superb statistics for a player who is not an out-and-out forward.
Ziyech will be looking to help Ajax extend a six-match unbeaten run against Feyenoord which stretches back over three years.
The Amsterdam side is in cracking form, having won three matches on the bounce (which includes 10 unanswered goals).
Erik ten Hag’s team defeated Benfica 1-0 in midweek to go top of the UEFA Champions League group, while their last Eredivisie outing on October 20 saw them rack up a 4-0 away win over Heerenveen.
Feyenoord, after a stuttering start to their season, have found consistency and confidence under manager Giovanni van Bronckhorst. Last weekend they won 3-0 at home to PEC Zwolle to extend their unbeaten run across all competitions to 10 matches.
In head-to-head terms, Ajax and Feyenoord have clashed in 157 Eredivisie matches dating back to 1921. De Godenzonen have claimed 69 wins compared to 44 for De Stadionclub, while 44 games have been drawn.