The area of square $A B C D$ is $300 \mathrm{~cm}^{2}$. In the figure, $M$ is the midpoint of $D C$ 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 $\triangle A F D$?

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-029.jpg?height=317&width=211&top_left_y=1235&top_left_x=1668)
(a) Note that $F \widehat{M} C$ and $A \widehat{M} D$ are vertically opposite angles, so $F \widehat{M} C = A \widehat{M} D$. Since $M C = M D$ and the triangles $\triangle A M D$ and $\triangle F M C$ are right triangles, we establish 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 was given, measuring $300 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-096.jpg?height=319&width=214&top_left_y=1094&top_left_x=1635)

(b) Since $A D = F C$ (from the previous item) and $M C = M D$, it follows that the triangles $\triangle A M D$, $\triangle D M F$, and $\triangle F M C$ have the same area. On the other hand, the sum of the areas of the last two is half the area of the square. Therefore, the area of the triangle $\triangle A F D$ is half the area of the square $A B C D$. This area was given, measuring $300 \mathrm{~cm}^{2}$, so the area of the triangle $\triangle A F D$ measures $150 \mathrm{~cm}^{2}$.