Use the law of cosines on triangles ABC and ABD



Using it on ABC



which simplifies to



Using it on ABD



which simplifies to



Equating the two expressions for cos(B),




Solve that and get



The height of both triangles is the green line h = AE:














approx. 226.8920448

Edwin

The triangle ABC is isosceles, therefore, its altitude AE is the median, at the same time.


So, I introduce the variable y = x/2.


Then BE = EC = y;  CD = x = 2y;  ED = 3y;  BD = 4y.


Therefore,

    h^2 = 17^2 - y^2 = 27^2 - (3y)^2,   or

    17 - y^2 = 27^2 - 9y^2

    9y^2 - y^2 = 27^2 - 17^2

    8y^2       = (27-17)*(27+17) = 10*44 = 440

     y^2                                 = 440/8 = 55

     y                                   = .


Then   h^2 = 17^2 - 55 = 234  and  h =  =  = .


Now the area of the triangle ABD is 


     area =  =  = 2y*h =  =  = 226.89  (approx.)