Example 36 (1992 Japan Mathematical Olympiad) Let the points dividing the sides $BC$, $CA$, $AB$ of $\triangle ABC$ internally in the ratio $t:(1-t)$ be $P$, $Q$, $R$, respectively. Let the area of the triangle formed by the segments $AP$, $BQ$, $CR$ be $K$, and the area of $\triangle ABC$ be $L$. Find $\frac{K}{L}$ (expressed in terms of $t$).
From the problem, it is felt that line segments $AP$, $BQ$, and $CR$ can be moved to form a triangle whose area can be calculated, thus allowing us to find the ratio $\frac{K}{L}$.

As shown in Figure 1-13, translate line segment $CR$ to $C'A$, forming parallelogram $ARCC'$. If we connect $PC'$ and have $PC' = BQ$, the initial goal can be achieved. In fact, connecting $C'Q$ in $\triangle CC'Q$ and $\triangle ABC$, let $BC = a$, $AC = b$, $AB = c$. Given $CQ = tb$, $CC' = AR = tc$, and $\angle QCC' = \angle CAB$, we know $\triangle CC'Q \sim \triangle ABC$.
Therefore, $C'Q = ta$, which means $C'Q = PB$.
Furthermore, from the similarity of triangles, $\angle CC'Q = \angle ABP$, so $C'Q \parallel PB$.
Thus, $BPC'Q$ is a parallelogram. Hence, $C'P = QB$.
Therefore, the triangle with sides $AP$, $BQ$, and $CR$ is $\triangle APC'$.
$$
\text{Thus, } \begin{aligned}
K & = S_{\triangle APC'} = S_{\triangle ABC} + S_{\triangle CC'Q} - S_{\triangle ABP} - S_{\triangle RCC'} \\
& = L + S_{\triangle ARC} - S_{\triangle BP} - \frac{1}{2}(1-t) a \cdot tc \cdot \sin (\pi - B) \\
& = L + tL - tL - t(1-t)L \\
& = (t^2 - t + 1) L.
\end{aligned}
$$

Therefore, $\frac{K}{L} = t^2 - t + 1$ is the desired result.