154 Regular tetrahedron $P-A B C$ has a base edge length of $a$, and side edge length of $b . D$ is on the side edge $P A$. The minimum area of $\triangle D B C$ is $\qquad$ .
154 If $b \geqslant \frac{\sqrt{2}}{2} a$, the minimum value is $\frac{a^{2}}{4 b} \sqrt{3 b^{2}-a^{2}}$. Otherwise, it is $\frac{a}{4} \sqrt{4 b^{2}-a^{2}}$.
Let the midpoint of $BC$ be $M$. Consider the section $APM$. (Figure 14(a)).
Figure 14
If $\angle APM = 90^{\circ}$ (Figure 14(d)), then $MD$ is minimized when $D$ coincides with $P$, at which time
$$
\begin{array}{c}
PM=\sqrt{PH^{2}+MH^{2}}=\sqrt{PA^{2}-AH^{2}+MH^{2}}=\sqrt{b^{2}-\frac{1}{4} a^{2}} . \\
S_{\triangle DBC}=\frac{a}{2} \sqrt{b^{2}-\frac{1}{4} a^{2}}=\frac{a}{4} \sqrt{4 b^{2}-a^{2}} .
\end{array}
$$