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2018 NCAA Women’s Tournament: #3 UCLA Meets #14 American U. for the First Time

UCLA is 44-3 at home since 2015-16.

The Bruins’ Road to the Final Four starts at home today and goes through Kansas City.
Joe Piechowski

After a two-week break, the UCLA Women’s Basketball team begins play in the 2018 NCAA Women’s Tournament today against the #14-seed American University Eagles. This is the first-ever meeting between these teams.

In fact, the last time UCLA and American met in any sport was in the 1985 NCAA Men’s Soccer Championship Game, which UCLA won 1-0.

The Eagles made the tournament by virtue of winning the Patriot League championship for the second time in four years. They defeated Navy 58-49 to win the title this year.

In that game, American was led by junior Cecily Carl who scored 20 points. Senior Emily Kinneston added 14 points. Kinneston led the Patriot in scoring with an average of 15.8 ppg and assists with 4.3 apg.

Overall, the Eagles have been a good defensive team. They have held opponents to only 57.4 ppg.

Of course, with UCLA playing at home, the Bruins are almost guaranteed to win. When playing at Pauley, UCLA is 44-3 since 2015-2016. Today, the Bruins hope to make it 45-3.

The winner of this game will face the winner of the other game played at Pauley today between #6-seed Iowa and #11-seed Creighton on Monday.

This is your UCLA Bruins vs. American Eagles NCAA Women’s Tournament, First Round open thread.

Go Bruins!!!