1996–97 UEFA Champions League knockout stage explained

The knockout stage of the 1996–97 UEFA Champions League began on 5 March 1997 and ended with the final at the Olympiastadion in Munich on 28 May 1997. The top two teams from each of the four groups in the group stage competed in the knockout stage. The draw for the quarter-finals was performed before the start of the group stage, with the winners of each group played against the runners-up from another; Group A vs. Group B, and Group C vs. Group D, with the group winner hosting the second leg. For the semi-finals, the winners of each tie between teams from groups A and B played against the winners of the corresponding tie between teams from groups C and D.

Qualified teams

The knockout stage involved the eight teams which qualified as winners and runners-up of all four groups in the group stage.

Groupwidth=210Winners
(seeded in quarter-final draw)
width=210Runners-up
(unseeded in quarter-final draw)
A Auxerre Ajax
B Atlético Madrid Borussia Dortmund
C Juventus Manchester United
D Porto Rosenborg

Format

Each quarter-final and semi-final was played over two legs, with each team playing one leg at home; the team that scored the most goals over the two legs qualified for the following round. In the event that the two teams scored the same number of goals over the two legs, the team that scored more goals away from home qualified for the next round; if both teams scored the same number of away goals, matches would go to golden goal extra time and then a penalty shoot-out if the teams could not be separated after extra time.

Quarter-finals

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First leg

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Second leg

Borussia Dortmund won 4–1 on aggregate.----Ajax won 4–3 on aggregate.----Juventus won 3–1 on aggregate.----Manchester United won 4–0 on aggregate.

Semi-finals

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First leg

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Second leg

Borussia Dortmund won 2–0 on aggregate.----Juventus won 6–2 on aggregate.

Final

See main article: 1997 UEFA Champions League Final.