Australian Shepherd Wins Best in Show!
| Print | By | June 12, 2009 1:11 AM

Australian Shepherd Wins Best in Show!

Over 2,000 of the country's top show dogs traveled to Philadelphia Thanksgiving Day for the National Dog Show Presented by Purina.

The Australian Shepherd, a member of the herding group, was named Best of the Best this year - taking the title of Best in Show.

More than 150 breeds competed for Best of Breed, First in Group and the highest canine honor of Best in Show.

Other winners included the Gordon Setter in the Sporting Group, Doberman in the Working Group, Beagle in the Hound Group, Sealyham Terrier in the Terrier Group, Schiperke in the Non-Sporting Group, and the Pekingese winning the toy group.

The Kennel Club of Philadelphia hosted the show which is also endorsed by the American Kennel Club.

The show was broadcast nationwide on NBC on Thursday, November 22nd 2008.

Comments (0)

Hide Posted Comments

Talk Bubble Iconadd your comment

Advertisement

Petside: Get Started

Advertisement

Specials

Check out these deals picked by petside.com just for you!

Newsletter & Deals

Register now for Newsletters and Personal Tools.

Your Name: Your Email:

Your privacy is important to us.
Click here for the full policy.