A heroic dog from Virginia saves a drowning fawn and stands by its side.
During the early evening of June 2, Ralph Dorn searched the lake’s surface behind his house for his Goldendoodle, Harley. Then he spotted the pup about 200 feet from shore; the 6-year-old canine was swimming with another animal, which Dorn discovered was a tiny baby deer after a few seconds.
“Not sure how the fawn got out there, but Harley obviously didn’t ask why, he just jumped into action,” Dorn, 62, of Culpeper, Virginia, wrote in a popular Facebook post about the event that has been shared more than 250,000 times.
Harley paddled beside the fawn all the way to shore. Dorn encountered the animals on land and assisted the fawn in climbing a difficult cliff. Harley began gently licking the fawn’s body after removing the baby deer from the pond and placing her on the grass.
“Harley really didn’t want to leave the fawn,” Dorn tells PEOPLE. “He just kept interacting with it, licking it and caring for it.”
Soon after the fawn reached the sea, the fawn’s mother came on the lawn. Dorn pulled Harley inside their adjacent home after he spotted the mother deer. The doe moved away with her baby after Dorn and Harley had left.
But something was wrong the next morning as Dorn and his wife, Patricia, 64, were having coffee.
“Harley became agitated as he ran from window to window. “When I opened the front door, we could hear the fawn bleating,” Dorn wrote on Facebook.
“Harley ran into the tree line and found the fawn,” he wrote. “The little one stopped bleating, tail wagging, they touched noses, sniffed each other, and Harley came back to the house calmly with me.”
Dorn tells PEOPLE that after the brief reunion with Harley, the fawn settled down, and by the end of the day, the baby and its mother were gone. Dorn hasn’t seen the pair since he estimated the fawn was a few days old.