In triangle $ABC$, $AB=\sqrt{30}$, $AC=\sqrt{6}$, and $BC=\sqrt{15}$. There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$, and $\angle ADB$ is a right angle. Find the value of the ratio
\[\dfrac{[ADB]}{[ABC]}.\]In other words, compute the ratio of the area of triangle $ADB$ to the area of triangle $ABC.$
Let $E$ be the midpoint of $\overline{BC}.$

[asy]
unitsize(1 cm);

pair A, B, C, D, E;

A = (0,0);
B = (sqrt(30),0);
C = intersectionpoint(arc(A,sqrt(6),0,180),arc(B,sqrt(15),0,180));
D = intersectionpoint(A--interp(A,(B + C)/2,5),Circle((A + B)/2, abs(A - B)/2));
E = (B + C)/2;

draw(A--B--C--cycle);
draw(A--D--B);

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, N);
label("$E$", E, N);
[/asy]

Then $BE = CE = \frac{BC}{2} = \frac{\sqrt{15}}{2},$ so by Stewart's Theorem applied to median $\overline{AE}$ of triangle $ABC,$
\[6 \cdot \frac{\sqrt{15}}{2} + 30 \cdot \frac{\sqrt{15}}{2} = \sqrt{15} \left( AE^2 + \frac{\sqrt{15}}{2} \cdot \frac{\sqrt{15}}{2} \right).\]This leads to $AE = \frac{\sqrt{57}}{2}.$

Let $x = DE$ and $y = BD.$  Then by the Pythagorean Theorem applied to right triangles $BDE$ and $BDA,$
\begin{align*}
x^2 + y^2 &= \frac{15}{4}, \\
\left( x + \frac{\sqrt{57}}{2} \right)^2 + y^2 &= 30.
\end{align*}Subtracting these equations, we get
\[x \sqrt{57} + \frac{57}{4} = \frac{105}{4},\]so $x = \frac{4 \sqrt{57}}{19}.$

Now, we want $\frac{[ADB]}{[ABC]}.$  Since $E$ is the midpoint of $\overline{BC},$ $[ABC] = 2 [ABE],$ so
\[\frac{[ADB]}{2 [ABE]} = \frac{AD}{2AE} = \frac{\frac{\sqrt{57}}{2} + \frac{4 \sqrt{57}}{19}}{2 \cdot \frac{\sqrt{57}}{2}} = \boxed{\frac{27}{38}}.\]