In Virginia, a dog saved a fawn from drowning in a lake. What happened next was nothing short of heartwarming. However, Ralph Dorn, a resident of Virginia, was shocked earlier when he could not find his dog.
The man’s pet, a Goldendoodle named Harley, was nowhere to be found. After some searching, the missing dog was soon found, and the story behind it was just so sweet.
Ralph Dorn was a retired Marine Corps pilot staying in Virginia. In one of his social media posts, he stated that he found his dog 200 feet from the shore of the lake behind his house.
The pet dog had 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.
The dog kept protecting the little one even when Harley and the baby deer were back on land. Harley just did not want to leave the baby deer. He kept on licking the adorable baby.
The following day, Harley’s owner noticed that his dog was restless. When he looked out of his front door, he saw that his dog had joined the fawn once again. It was a beautiful reunion as the fawn had stopped by to thank his savior and then walked away into the forest.
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