2.31. The side of the base of a regular triangular pyramid is equal to $a$. A plane is drawn through one of the sides of the base, perpendicular to the opposite lateral edge and dividing this edge in the ratio $\boldsymbol{m}: \boldsymbol{n}$, counting from the vertex of the base. Determine the total surface area of the pyramid.
2.31. The full surface area of the pyramid can be found using the formula $S_{\text {full }}=\frac{a^{2} \sqrt{3}}{4}+\frac{1}{2} \cdot 3 a \cdot S D$ (Fig. 2.26). Since $\triangle B O S \sim \triangle B K D$ (right triangles with a common angle), we have

$$
\frac{B D}{B S}=\frac{B K}{B O}, \text { or } \frac{a \sqrt{3}}{2}: B S=B K: \frac{a \sqrt{3}}{3}
$$

i.e., $a^{2}=2 B K \cdot B S$. According to the condition,

$$
\frac{B K}{K S}=\frac{m}{n}, \text { hence }
$$

$$
K S=\frac{n}{m} B K, B S=B K+K S=\frac{m+n}{m} B K, B K=\frac{m}{m+n} B S
$$

Thus, $a^{2}=\frac{2 m}{m+n} B S^{2}$, i.e., $B S^{2}=\frac{(m+n) a^{2}}{2 m}$. In $\triangle S O D$, we have $S D^{2}=S O^{2}+O D^{2}$, hence, considering that $S O^{2}=B S^{2}-B O^{2}$, we find

$$
S D^{2}=\frac{(m+n) a^{2}}{2 m}-\frac{a^{2}}{3}+\frac{a^{2}}{12}=\frac{3 m a^{2}+6 n a^{2}}{12 m}=\frac{(m+2 n) a^{2}}{4 m}
$$

Thus,

$$
\begin{aligned}
& S_{\text {full }}=\frac{a^{2} \sqrt{3}}{4}+\frac{1}{2} \cdot 3 a \cdot \frac{a}{2} \sqrt{\frac{m+2 n}{m}}=\frac{a^{2} \sqrt{3}}{4}\left(1+\sqrt{\frac{3(m+2 n)}{m}}\right) \\
& \text { Answer: }: \frac{a^{2} \sqrt{3}}{4}\left(1+\sqrt{\frac{3(m+2 n)}{m}}\right)
\end{aligned}
$$