Thanks, Mr, Sud; I agree; but you've been of no help
I had solved it (also with a darn Quartic!) and was sneakily trying
to see if someone could come up with something "simpler".
Since I hate using SIN or COS, I placed CB on x-axis with C at origin;
let h = AH, so A(56,h).
Letting (x,y) = D's coordinates, I used following equations:
BD's y-intercept is clearly 84; hence BD's equation is: y = (-3/4)x + 84
AC's equation is easier still, with points (0,0) and (56,h) : y = (h/56)x
So I needed to solve:
(56 - x)^2 + (h - y)^2 = 51^2
Getting x and y in terms of h:
x = 4704 / (h + 42)
y = 84h / (h + 42)
And that leads to MY(!) quartic:
h^4 - 84h^3 + 2299h^2 - 481908h + 943740 = 0 (hope it's better than yours)
Which has h = 105 as only "valid" solution (so AE = 63)
And that checks out ok. Makes the equal sides AB and AC = 119 ; also DE = 30.
Surprising to me that this is not easier, with right triangles all over the place!