Keuchel was a two-time All-Star in seven seasons with Houston. 

By Michael Shapiro
June 06, 2019

At long last, hot stove season is over. Dallas Keuchel is going to Atlanta.

Keuchel, the former Astros lefthander and the 2015 Cy Young Award winner, has agreed to a one-year deal with the Braves, the team announced on Friday. The Athletic's David O'Brien first reported on Thursday that Keuchel was signing with Atlanta. Tim Brown of Yahoo! Sports was first to report the terms of the deal, which is for $13 million. Per Brown, Keuchel is scheduled to start on Saturday for Triple-A Gwinnett against Durham.

The $13 million in the deal is prorated and would be worth around $20 million for a full season, according to ESPN's Jeff Passan. Interestingly, according to ESPN's Buster Olney, the Astros offered Keuchel two possible deals in March: $15 million for one year and a two-year/$24 million contract.

The 31-year-old spent his first seven seasons in Houston, where he was a two-time All-Star.

Keuchel finished the 2018 season with a 3.74 ERA, striking out 153 batters in 204 2/3 innings. He led the American League with 34 starts, but he also allowed 211 hits, the most in the majors. In his Cy Young campaign, Keuchel went 20–8 with a 2.48 ERA and also led MLB with 232 innings pitched.

Despite being one of the marquee free agents this winter, Keuchel began the 2019 campaign without a contract and spent much of the first two months of the season working out and throwing simulated games, waiting for offers. Many teams didn't want to sign either Keuchel or closer Craig Kimbrel until after June 3, when they would no longer have to forfeit a pick in the 2019 MLB Draft as compensation. Keuchel was expected to be signed soon after his free agency became unrestricted, and the Yankees and Braves emerged as favorites to sign the lefty.

Keuchel won four Gold Gloves with the Astros. He brings an impressive playoff resume to Atlanta, tallying a 3.31 ERA in nine postseason starts since 2015.

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