NEWS BULLETIN: your intuitive 90 degrees was CORRECT!!
Let h = height of triangle; place the triangle on x-axis:
(I'm using "b" as half the triangle's base; FORGET the OP's "b"!)
We know that b = sqrt(6)/2 = ~1.224745
C(0,0) H(b,0) B(2b,0)
Since angle EBH = 15, then a = bSIN(15)/SIN(75) = ~.328169
Equation of AC: y = (h/b)x 
Equation of BD: y = (-a/b)x + 2a 
So, using  and :
x = 2ab / (a + h)
y = 2ah / (a + h)
So, using AD as hypotenuse, we have:
(b - x)^2 + (h - y)^2 = 1^2
Substituting for x and y leads to:
h^4 - (2a)h^3 + (a^2 + b^2 + 1)h^2 - (2ab^2 + 2a)h + a^2(b^2 - 1) = 0
Using values to accuracy .000000:
h^4 - .656338h^3 + .607695h^2 - 1.640847h + .053847 = 0
Thanks to Wolfram: h = 1.224745
Let e = AB: e = sqrt(b^2 + h^2)
So angle BAH = ASIN(b/e) = 45 degrees !!
(my degree of accuracy results in 44.999996991.....)
Plus Sir Skeeter's warning that this wouldn't be short/simple was also correct!
EDIT: Note that triangle ADB = 30-60-90, thus the OP's b (line BD) = 2.