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"!)

Code:

A(b,h)
(1)
D(x,y)
E(b,a)
C(0,0) H(b,0) B(2b,0)

We know that b = sqrt(6)/2 = ~1.224745

Since angle EBH = 15, then a = bSIN(15)/SIN(75) = ~.328169

Equation of AC: y = (h/b)x [1]

Equation of BD: y = (-a/b)x + 2a [2]

So, using [1] and [2]:

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.