We will be using the law of cosines.
Since the adjacent interior angles of a parallogram are
supplementary, cos(∠ABC) = -cos(∠DAB), so

Using the law of cosines on ΔABD

       BD² =  AD² +  AB² - 2∙ AD∙ AB∙cos(∠DAB)
(1)   602² = 418² + 694² - 2∙418∙694∙cos(∠DAB)  

Using the law of cosines on ΔABC

       AC² =  BC² +  AB² - 2∙ BC ∙AB∙cos(∠ABC)
       AC² = 418² + 694² - 2∙418∙694∙[-cos(∠DAB)]
(2)    AC² = 418² + 694² + 2∙418∙694∙cos(∠DAB)

Adding equations (1) and (2)

(1)   602² = 418² + 694² - 2∙418∙694∙cos(∠DAB)
(2)    AC² = 418² + 694² + 2∙418∙694∙cos(∠DAB)

602² + AC² = 2∙418² + 2∙694² 
       AC² = 2∙418² + 2∙694² - 602²
       AC² = 2∙418² + 2∙694² - 602²
       AC² = 349448 + 963272 - 362484
       AC² = 950316
        AC = √950316
        AC = 974.8415256 ft.   

Edwin