The given figure represents a trapezoid $A B C D$ where $A B$ is parallel to $C D$ and the diagonals $A C$ and $B D$ intersect at point $P$. If the areas of triangles $\triangle A P B$ and $\triangle C P D$ are 4 and $9 \mathrm{~cm}^{2}$, respectively, what is the area of triangle $\triangle P C B$?

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-027.jpg?height=257&width=357&top_left_y=1682&top_left_x=1523)
The triangles $\triangle A P B$ and $\triangle C P D$ are similar because the angle $A \widehat{P} B$ is equal to the angle $C \widehat{P} D$ (vertical angles) and the angle $A \widehat{B} D$ is equal to the angle $B \widehat{D} C$ (alternate interior angles).

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-090.jpg?height=360&width=537&top_left_y=1770&top_left_x=842)

Since the ratio of their areas is $4 / 9$, the similarity ratio between these triangles is $\sqrt{4 / 9}=2 / 3$. Therefore,

$$
\frac{P B}{D P}=\frac{2}{3}
$$

On the other hand, the triangles $\triangle C P D$ and $\triangle P C B$ have the same height relative to the bases $D P$ and $P B$, respectively. Therefore, the ratio of their areas is equal to the ratio of their bases,

$$
\frac{\text { area }(\triangle P C B)}{\text { area }(\triangle C P D)}=\frac{P B}{D P}=\frac{2}{3}
$$

Since the area of $(\triangle C P D)=9$, it follows that the area of the triangle $\triangle P C B$ measures $6 \mathrm{~cm}^{2}$.

Note: The same argument could be used to show that the area of the triangle $\triangle A D P$ also measures $6 \mathrm{~cm}^{2}$. The area measures of 4 and $9 \mathrm{~cm}^{2}$ given in the problem statement do not play any special role. The argument presented above proves that in any trapezoid, the triangles $\triangle P C B$ and $\triangle A D P$ have equal areas, even if the trapezoid is not isosceles.