2. In the tetrahedron $S-ABC$, $SA \perp$ plane $ABC$, $AB \perp BC$, point $E$ is the midpoint of $SC$, $D$ is a point on $AC$, and $DE \perp SC$. Also, $SA=AB$, $SB=BC$, find the dihedral angle with $BD$ as the edge and $BDE$ and $BDC$ as the faces.
2. From the given conditions, $E$ is the midpoint of $SC$, and $SB=BC$, so we know that $BE \perp SC$. Combined with $DE \perp SC$, it follows that $SC \perp$ plane $BDE$. This indicates that $\triangle BDE$ is the projection of $\triangle BDC$ on plane $BDE$. Let the angle of the dihedral angle be $\alpha$, then by property 8, we have $S_{\triangle BDE} = S_{\triangle BDC} \cdot \cos \alpha$.

Since $SC \perp$ plane $BDE$ and $SE = EC$, we know that $V_{S-BDE} = V_{C-BDE}$, i.e., $V_{C-BDE} = \frac{1}{3} SE \cdot S_{\triangle BDE}$. Noting that $SA \perp$ plane $ABC$ and $E$ is the midpoint of $SC$, the height from point $E$ to plane $BCD$ is equal to $\frac{1}{2} SA$. Therefore, $V_{E-BCD} = \frac{1}{2} \cdot \frac{1}{2} SA \cdot S_{\triangle BDC}$, and since $V_{C-BDE} = V_{E-BCD}$, we have $\frac{1}{3} SE \cdot S_{\triangle BDE} = \frac{1}{6} \cdot SA \cdot S_{\triangle BDC}$, hence $\cos \alpha = \frac{S_{\triangle BDE}}{S_{\triangle BDC}} = \frac{SA}{2SE} = \frac{SA}{SC}$.

Given that $SA \perp$ plane $ABC$, it follows that $SA \perp BC$. Combined with $BC \perp AB$, we know that $BC \perp$ plane $SAB$, thus $BC \perp AB$, so $SC = \sqrt{2} SB = \sqrt{2}(\sqrt{2} SA) = 2SA$, hence $\cos \alpha = \frac{1}{2}$, i.e., $\alpha = 30^\circ$.