1. Let the points dividing the sides $BC, CA, AB$ of $\triangle ABC$ internally in the ratio $t(1-t)$ be $P, Q, R$, respectively. The area of the triangle formed by the segments $AP, BQ, CR$ is $K$, and the area of $\triangle ABC$ is $L$. Find $\frac{K}{L}$ (expressed in terms of $t$).
1. As shown in the figure, regarding line $AB$, take $C^{\prime}$ on the same side as $C$, such that $A R C C^{\prime}$ forms a parallelogram. In $\triangle C C^{\prime} Q$ and $\triangle A B C$,
$$
\begin{array}{l}
\because C Q=t b, C C^{\prime}=t c, \\
\angle Q C C^{\prime}=\angle C A B, \\
\therefore \triangle C C^{\prime} Q \sim \triangle A B C .
\end{array}
$$

Therefore, $C^{\prime} Q=t a$, i.e., $C^{\prime} Q=P B$.
$$
\because \angle C C^{\prime} Q=\angle A B P \text {, }
$$
$\therefore B P C^{\prime} Q$ is a parallelogram.
Thus, the triangle formed by $A P, B Q, C R$ is $\triangle A P C^{\prime}$. Therefore,
$$
\begin{aligned}
K & =\triangle A P C^{\prime} \\
& =L+\triangle A C C^{\prime}-\triangle A B P-\triangle P C C^{\prime} \\
& =L+\triangle A R C-\triangle A B P-\frac{1}{2}(1-t) a
\end{aligned}
$$
$$
\text { - } \begin{aligned}
t c & =\sin (\pi-B) \\
& =L+t L-t L-t(1-t) L \\
& =\left(t^{2}-t+1\right) L, \\
\therefore & \frac{K}{L}=t^{2}-t+1 .
\end{aligned}
$$