Let's use some coordinate geometry here and WLOG, let $a=1$...let the hypotenuse lie along the line:
$$y=1-\frac{x}{b}$$
And line segment $\overline{CD}$ lie along the line:
$$y=\cot(3u)x$$
And so we find $D$ is at:
$$(x,y)=\left(\frac{b}{b\cot(3u)+1},\frac{b\cot(3u)}{b\cot(3u)+1}\right)$$
Now, we require:
$$(b-1)^2=\left(\frac{b}{b\cot(3u)+1}\right)^2+\left(\frac{1}{b\cot(3u)+1}\right)^2$$
From this, we determine:
$$\cot(3u)=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}$$
And using a triple-angle identity for cotangent, we have:
$$\frac{3\cot(u)-\cot^3(u)}{1-3\cot^2(u)}=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}$$
We also know:
$$\cot(2u)=b$$
and using a double-angle identity for cotangent, we have:
$$\frac{\cot^2(u)-1}{2\cot(u)}=b$$
Form this, we determine:
$$\cot(u)=b+\sqrt{b^2+1}$$
And so we have:
$$\frac{\left(b+\sqrt{b^2+1}\right)\left(b^2+b\sqrt{b^2+1}-1\right)}{3b^2+3b\sqrt{b^2+1}+1}=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}$$
Taking the positive real root, we find:
$$b=\sqrt{3}$$
And this in turn implies:
$$\cot(3u)=1$$
or:
$$3u=45^{\circ}$$
$$u=15^{\circ}$$
