9 Times Tina Fey & Amy Poehler Slayed the Red Carpet Together
Since we first saw their undeniable chemistry (and matching Weekend Update blazers) on Saturday Night Live, we’ve been obsessed with everything Tina Fey and Amy Poehler. And just like how they complement each other in comedy, the longtime BFFs have also found their groove on the red carpet.
Fey, who works with stylist Cristina Ehrlich, usually favors classic, feminine shapes by designers like Oscar de la Renta, which is a nice contrast to Poehler’s more minimalist aesthetic (Stella McCartney is a fave). “I collaborate with Amy’s stylist Karla Welch the same way that Tina collaborates with Amy on projects,” Ehrlich told us earlier this year. "Sometimes what Tina wears depends on what Amy chooses, so as stylists we support each other and always keep each other in the loop."
But that doesn’t mean that the stars will always look dramatically different at events. Just last week, they were twinning in sheer black gowns at the premiere of their latest movie, Sisters. Check out that look and all of the other times they’ve slayed the red carpet as a dynamic duo, below.
The New York Premiere of Sisters (2015)
The Golden Globes (2015)
The Golden Globes (2014)
The Screen Actors Guild Awards (2013)
Wonder if they planned this? The stars, who were both nominated for Outstanding Performance by a Female Actor in a Comedy Series, donned similar corsetted black gowns, drop earrings, and bouncy blow outs. Fey chose Oscar de la Renta, while Poehler fancied Zuhair Murad.
The Golden Globes (2013)
The Warner Bros. & InStyle Golden Globes After Party (2012)
The Screen Actors Guild Awards (2011)
The New York Premiere of Baby Mama
To fête their comedy, the besties wore tiered cocktail dresses. Fey's had a sophisticated Audrey Hepburn vibe, while Poehler's was light and airy.
The Women of the Year Awards (2002)
The gang's all here! The besties, in formfitting cocktail dresses and heels, looked gorgeous as they joked around with SNL castmates Rachel Dratch, Ana Gasteyer, and Maya Rudolph.