USA dominates Ryder Cup in bizarre foursome finish

Phil Mickelson and Keegan Bradley of the United States steamrolled Europe's Lee Westwood and Luke Donald 7 and 6 in the second of Saturday's morning foursomes at the Ryder Cup. The win by the two former major championship winners ties the record for the most one-sided victory in Ryder Cup team play history. Mickelson, who is playing in his 37th Ryder Cup match which ties Billy Casper for the most by a US player, needed another magical shot on the final No. 12 hole after after an errant tee shot. Rookie Bradley aggressively got it out of the trees and then Mickelson's approach landed 20 feet away, spun left and carried down the slope to within a foot of the cup. Westwood missed a birdie putt and Donald failed to convert his par putt as the Americans routed the European duo at the Medinah Country Club course. Mickelson and Bradley got off to another hot start by making birdie on the first two holes and they never took their foot off the gas. They closed it out in dramatic fashion by winning five of the final seven holes. The two hit it off together in fine style on Friday, winning in both foursomes and fourballs to ignite the US charge that resulted in a 5-3 overnight lead. On the par-four opening hole, Mickelson hit a fantastic approach leaving it about a foot from the cup for an easy birdie which Europe conceded. Donald missed his seven footer and the first point went to the Americans. Mickelson hit another great shot on the par-three second as his tee shot landed about 12 feet away while Westwood pulled his drive 20 yards left into the water and the US was off to the races. Europe's putting woes continued on the difficult par-four sixth as Donald lipped out an eight footer and the Americans went 3-up. Team USA posted back-to-back birdies as Bradley just missed his eagle attempt, giving Mickelson a short putt to win the hole and a 4-up lead through seven.