## Task B-1.7.

In triangle $A B C$, point $D$ is on side $\overline{B C}$, and point $E$ is on side $\overline{A C}$. Segments $\overline{B E}$ and $\overline{A D}$ intersect at point $F$, and segment $\overline{B F}$ divides the area of triangle $A B D$ into two equal parts. If the area of triangle $A F E$ is 7, and the area of triangle $A F B$ is 13, calculate the areas of triangles $C E F$ and $C F D$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_3d5a33db4db92485fe9eg-05.jpg?height=617&width=720&top_left_y=594&top_left_x=631)

Let $x$ be the area of triangle $CEF$, and $y$ be the area of triangle $CFD$.

Triangles $AFB$ and $FBD$ have a common height from vertex $B$, and since they have the same areas, they also have the same base lengths to which the height is dropped.

1 point

Therefore, $|AF| = |FD|$.

1 point

This means that triangles $AFC$ and $FDC$ have the same bases to which the common height from vertex $C$ is dropped, so they have the same areas, i.e.:

$$
y = x + 7
$$

(*) 2 points

Notice that triangles $AFE$ and $ABF$ have a common height from vertex $A$, which means the ratio of their areas is equal to the ratio of the lengths of the bases $\overline{EF}$ and $\overline{FB}$:

$$
\frac{|EF|}{|FB|} = \frac{7}{13}
$$

On the other hand, triangles $EFC$ and $FBC$ have a common height to the same bases, so the ratio of their areas is equal to the ratio of the lengths of those bases. We have:

$$
\frac{x}{y + 13} = \frac{7}{13}
$$

which together with (*) gives

$$
\frac{x}{x + 20} = \frac{7}{13}
$$

1 point

i.e., $13x = 7x + 140 \Rightarrow 6x = 140$, so the areas are $x = \frac{70}{3}$ and $y = \frac{70}{3} + 7 = \frac{91}{3}$.

1 point

## SCHOOL/CITY COMPETITION IN MATHEMATICS

2nd grade - high school - B variant
March 3, 2021.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.