CINCINNATI, OH (FOX19) - The American Institute of Plastic Surgeons has recognized the exceptional performance of Ohio’s Plastic Surgeon Jon E. Mendelsohn, M.D. as among the 2018 10 Best Plastic Surgeons for Client Satisfaction.
The American Institute of Plastic Surgeons is a third-party rating organization that publishes an annual list of the Top 10 Plastic Surgeon in each state. Surgeons who are selected to the “10 Best” list must pass American Institute of Plastic Surgeons’ rigorous selection process, which is based on patient and/or peer nominations, thorough research, and American Institute of Plastic Surgeons’ independent evaluation.
American Institute of Plastic Surgeons’ annual list was created to be used as a resource for patients during the surgeon selection process. One of the most significant aspects of the selection process involves surgeons' relationships and reputation among his or her patients.
As patients should be a surgeon’s top priority, American Institute of Plastic Surgeon places the utmost emphasis on selecting surgeons who have achieved significant success in the field of Plastic Surgery without sacrificing the service and support they provide. Selection criteria therefore focus on surgeons who demonstrate the highest standards of Client Satisfaction.
The institute congratulates Jon E. Mendelsohn, M.D. on this achievement and said it is honored to have him as a 2018 American Institute of Plastic Surgeons’ Member.
More information on Mendelsohn’s practice is available at his website.