A very sadistic geometry question?

Let's label the triangle A, B, C with A being the right angle.

D = incenter

E = midpoint of BC

angle ABC = θ

BC = 4, BE = CE = 2

AD = r√2

AE = 2

where r = radius of incircle

angle EAD = 45 - θ

So far area of EAD = 1/2 * AD * AE * sin(EAD)

= r*(cosθ - sinθ)

Let F be the point along AC where the incircle intersects AC.

Triangle DFC is right angled, such that

angle DCF = 45 - θ/2

DF = r, CF = 4sinθ - r

r / (4sinθ - r) = tan(45-θ/2) = (1 - tanθ/2) / (1+tanθ/2)

From this

r = 2 sinθ * (1 - tanθ/2)

and area

A = 2sinθ*(cosθ - sinθ) * (1 - tanθ/2)

Now we need to set dA/dθ = 0 and solve.

At this point, there is a way to simplify the half angle part.

tan(x) = 2tan(x/2) / (1 - tan^2 (x/2))

From this we can obtain

tan(x/2) = (1 - cosx) / sinx

1 - tan(x/2) = (sinx + cosx - 1) / sinx

Now A = 2*(cos2θ + sinθ - cosθ)

dA/dθ = 2*(-2sin2θ + cosθ + sinθ) = 0

Of course you could set y = sinθ, then end up with a quartic equation in y. But I'm looking for another trick that can help us solve this. So far no luck. A numerical solution yields θ = 19.91 degrees.

The quartic equation in y = sinθ is

16y^4 - 8y^3 - 14y^2 + 8y - 1 = 0

*EDIT*

OK thanks to Duke for a little help on this one.

cosθ + sinθ = 2sin2θ

(cosθ + sinθ)^2 = 4(sin2θ)^2

1 + sin2θ = 4(sin2θ)^2

sin2θ = (1 + √17)/8

θ = (1/2)*arcsin[(1 + √17)/8]

let the hypotenuse lie on x-axis, from (0,0) to (4,0).

also the smallest angle be x, situated at (4,0).

right-angled vertex, RAV's coordinate

= ((4/2)(1-cos2x) , (4/2)sin2x)

= (2-2cos2x , 2sin2x)

side opposite (0,0)'s length

= 4cosx

side opposite (4,0)'s length

= 4sinx

incenter's coordinate

= ([4*4sinx+4(2-2cos2x)]/[4(sinx+cosx+1)] , [4*2sin2x]/[4(sinx+cosx+1)])

= ([4sinx+2-2cos2x)]/[sinx+cosx+1] , [2sin2x]/[sinx+cosx+1])

= (4sinx(sinx+1)/[sinx+cosx+1] , 4sinx(cosx)/[sinx+cosx+1])