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Two wins put Ferrer in Auckland finals

David Ferrer, shown in a tournament last November, won a pair of matches Friday and will play Saturday in the title match of the Heineken Open tennis tournament in New Zealand. UPI/David Silpa
David Ferrer, shown in a tournament last November, won a pair of matches Friday and will play Saturday in the title match of the Heineken Open tennis tournament in New Zealand. UPI/David Silpa | License Photo

AUCKLAND, New Zealand, Jan. 13 (UPI) -- No. 1-seed and defending champion David Ferrer won twice Friday and is returning to the finals of the Heineken Open tennis tournament in New Zealand.

Ferrer, No. 5 in the world, was forced into double duty when rain washed out his quarterfinal match Thursday. He battled Alejandro Falla for more than 2 hours for a 6-4, 4-6, 6-3 win and then beat third-seeded Fernando Verdasco 6-3, 6-4 in the semifinals.

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Verdasco, who also had to finish his quarterfinal Friday, didn't manage a break point against Ferrer. Ferrer won three-quarters of the points on serve and had 12 break opportunities of his own. He finished off one in each set.

He'll next go against Olivier Rochus, who upset eighth-seeded Philipp Kohlschreiber 6-7 (4-7), 6-1, 6-4 Friday. Rochus had six breaks -- two in each set -- in earning his 10th ATP finals appearance. He's 2-7 in those matches with his most recent title coming in 2006.

Ferrer advanced to his 26th title match. He was 2-4 in finals last year.

Ferrer is 5-2 lifetime against Rochus but they haven't played since 2008.

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