8、In tetrahedron $A B C D$, it is known that $D A \perp$ base $A B C$, side $A B D \perp$ side $B C D, B D=B C=2$, the sum of the squares of the areas of the three sides $D A B, D B C, D C A$ is 8, then $\angle A D B=$ $\qquad$
8. $\frac{\pi}{4}$

Analysis: Draw a perpendicular line from point $A$ to $BD$ intersecting at point $E$. Since the side $ABD \perp$ side $BCD$, it follows that $AE \perp$ plane $BCD$, i.e., $AE \perp BC$. Because $AD \perp BC$, it follows that $AB \perp BC$; Let $\angle ADB = \theta$, then $AD = 2 \cos \theta, AB = 2 \sin \theta, AC = 2 \sqrt{\sin^2 \theta + 1}$. Therefore, $8 = \left(2 \cos \theta \sqrt{\sin^2 \theta + 1}\right)^2 + 4 + (2 \cos \theta \sin \theta)^2$, which simplifies to $\cos \theta = \frac{\sqrt{2}}{2}$, i.e., $\theta = \frac{\pi}{4}$;