6. As shown in the figure, $D$ is a point on side $B C$, and $B D=2 D C, D P / / C A$. The area of triangle $A P D$ is $14 \mathrm{~cm}^{2}$, what is the area of triangle $A B C$ in $\mathrm{cm}^{2}$.
6. As shown in the figure, $D$ is a point on side $BC$, and $BD=2DC$, $DP \parallel CA$. The area of triangle $APD$ is $14 \text{ cm}^2$. What is the area of triangle $ABC$ in $\text{cm}^2$?

Connect $PC$. Since $DP \parallel CA$, and the areas of triangles with equal bases and heights are equal, $S_{\triangle PCD}=S_{\triangle APD}=14$. Also, because $BD=2DC$, the area ratio of two triangles with the same height is equal to the ratio of their base lengths. Therefore, $S_{\triangle PBD}$ is 2 times $S_{\triangle PCD}$, from which we can find the area of $S_{\triangle ABD}$, and then the area of triangle $ABC$.

【Solution】Solution: Connect $PC$, see the figure above,
Since $DP \parallel CA$,
Therefore, $S_{\triangle PCD}=S_{\triangle APD}=14$,
Also, since $BD=2DC$,
Therefore, $S_{\triangle PBD}=2 \times 14=28$ (square centimeters),
Therefore, $S_{\triangle ABD}=S_{\triangle PBD}+S_{\triangle APD}=28+14=42$ (square centimeters),
Therefore, $S_{\triangle ABC}=\frac{3}{2} \times S_{\triangle ABD}=\frac{3}{2} \times 42=63$ (square centimeters),
Answer: The area of triangle $ABC$ is 63 square centimeters.