Former Cleveland Browns safety Juan Thornhill agreed to terms with the Pittsburgh Steelers on a one-year contract Monday night.

The deal was first reported by Ian Rapoport of NFL Network on social media.

Thornhill was deciding between the Steelers and the San Francisco 49ers, whom he visited last Thursday. In the end, Thornhill believed the Steelers gave him the best opportunity.

Originally drafted by the Kansas City Chiefs in the second round of the 2019 NFL Draft out of Virginia, Thornhill played for the Chiefs for four years and became an unrestricted free agent.

Thornhill then signed with the Browns in free agency, who inked him to a three-year deal worth $21 million. He was released by the Browns in February heading into the final year of his deal, which was worth $7 million.

It is believed he was released due to injury concerns as he missed six games each of the last two seasons. He also had a high cap number for a player who missed that amount of action.

Thornhill appeared in 11 games for Cleveland in 2024. He totaled 49 tackles and had three passes broken up. For his career, Thornhill has amassed eight interceptions and returned one for a touchdown his rookie season. He has appeared in 87 games and started 74 of them.

He will more than likely backup Minkah Fitzpatrick at the free safety position and come in on nickel packages. DeShon Elliott has a lock on the strong safety position and played well last season.

By signing with the Steelers, Thornhill will have two chances to show the Browns they made a mistake in releasing him. The Browns and the Steelers, both residents of the AFC North, face off twice a year.

Terms of the deal have not been disclosed.