The area of square $A B C D$ is $300 \mathrm{~cm}^{2}$. In the figure, $M$ is the midpoint of $C D$ and point $F$ lies on the line passing through $B$ and $C$.

(a) What is the area of triangle $\triangle A B F$?

(b) What is the area of triangle $\triangle A D F$?

![](https://cdn.mathpix.com/cropped/2024_05_01_eead07372ad548e635bag-03.jpg?height=358&width=225&top_left_y=680&top_left_x=1489)
(a) Note that $F \widehat{M} C$ and $A \widehat{M} D$ are vertically opposite angles. Therefore, $F \widehat{M} C = A \widehat{M} D$. Since $M C = M D$ and the triangles $\triangle A M D$ and $\triangle F M C$ are right triangles, we conclude that they are congruent. Thus, they have the same area, from which we conclude that the area of the triangle $\triangle A B F$ is equal to the area of the square $A B C D$, which is $300 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_eead07372ad548e635bag-17.jpg?height=396&width=260&top_left_y=688&top_left_x=1366)

(b) Since $A D = F C$ (from the previous item) and $D M = M C$, it follows that the triangles $\triangle A D M$, $\triangle D M F$, and $\triangle M C F$ have the same area. On the other hand, the area of the last two is half the area of the square. Therefore, the area of the triangle $\triangle A D F$ is half the area of the square, which is $150 \mathrm{~cm}^{2}$.