A dog in Virginia saved a fawn from drowning in a lake. What followed was nothing short of heartwarming. Ralph Dorn, a Virginia resident, was shocked earlier when he couldn’t find his dog.
Harley, the man’s Goldendoodle, was nowhere to be found. The missing dog was soon found after some searching, and the story behind it was simply adorable.
Ralph Dorn was a former Marine Corps pilot who lived in Virginia. He stated on social media that he found his dog 200 feet from the shore of the lake behind his house.
The pet dog realized that he was not alone as he saw a baby deer fighting to get back to the shore of the lake. The brave dog had jumped into the water to save the little fawn.
Harley and the baby deer soon reached the shoreline safely. The dog’s owner said that he was proud of his brave dog. However, the story did not just end there for the two animals.
Even after Harley and the baby deer returned to land, the dog continued to protect the youngster. Harley simply did not want to abandon the baby deer. He kept licking the cute baby.
Harley’s owner noticed his dog was restless the next day. When he looked out his front door, he noticed that his dog had rejoined the fawn. The fawn had stopped by to thank his savior before walking away into the forest, making it a beautiful reunion.