Former WWE star responds to Danhausen
Former WWE star responds to Danhausen Notifications New User posted their first comment this is comment text Approve Reject & ban Delete Logout
Yes, please. Stealing is wrong. That wouldn’t be a babyface thing to do. I’m a role model to the youth of America. Yes, please. FTR is currently enjoying an overwhelmingly well-received run as babyfaces. The duo initially broke into AEW as one of the top heel factions before having a change of heart a few months ago. As one of the top tag teams in the promotion, could be the first two-time AEW Tag Team Champions? Need to catch up with the latest AEW Dynamite results? Check them out via .
Stealing is wrong – Former WWE star responds to interesting proposition from Danhausen
Has Danhausen irked this former WWE Superstar? WWE was once home to Dax Harwood (FKA Scott Dawson), but these days the one-half of the ROH Tag Team Champion seems more comfortable in AEW. While FTR is still gunning for the AEW Tag Titles, Danhausen recently offered to steal the belts for Dax. The Young Bucks recently they'll be capturing the AEW tag titles this coming Wednesday. Harwood, who recently defeated the Bucks alongside Cash Wheeler, took offense and . While the former WWE Superstar initially seemed to have declined, he jokingly took Danhausen's offer: "Stealing is wrong. That wouldn’t be a babyface thing to do. I’m a role model to the youth of America. Yes, please." - Dax Tweeted. Stealing is wrong. That wouldn’t be a babyface thing to do. I’m a role model to the youth of America.Yes, please. Stealing is wrong. That wouldn’t be a babyface thing to do. I’m a role model to the youth of America. Yes, please. FTR is currently enjoying an overwhelmingly well-received run as babyfaces. The duo initially broke into AEW as one of the top heel factions before having a change of heart a few months ago. As one of the top tag teams in the promotion, could be the first two-time AEW Tag Team Champions? Need to catch up with the latest AEW Dynamite results? Check them out via .