8,9 $[\quad$ The Law of Cosines

In an equilateral triangle $ABC$ with side $a$, points $E$ and $D$ are the midpoints of sides $BC$ and $AC$ respectively. Point $F$ lies on segment $DC$, and segments $BF$ and $DE$ intersect at point $M$. Find $ME$, given that the area of quadrilateral $ABMD$ is $\frac{5}{8}$ of the area of triangle $ABC$.

#
$D M=\frac{a}{4}, D F=\frac{a}{6}$.

## Solution

The height of the trapezoid $A B M D$ is half the height of the triangle $A B C$. Therefore,

$$
S_{\mathrm{ABMD}}=\frac{1}{2}(A B+M D) h=\frac{5}{8} \cdot \frac{1}{2} A B \cdot h,
$$

where $h$ is the height of the triangle $A B C$. From this, we find that $M D=\frac{1}{4} A B=\frac{a}{4}$. From the similarity of triangles $M D F$ and $B A F$, it follows that $D F=\frac{1}{3} A D=\frac{a}{6}$. From the triangle $D M F$, we find by the cosine theorem that

$$
M F=\sqrt{D F^{2}+D M^{2}-2 D F \cdot D M \cos 60^{\circ}}=
$$

$$
=a \sqrt{\frac{1}{36}+\frac{1}{16}-\frac{1}{24}}=\frac{a \sqrt{7}}{12}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_65daf5394c369f17193bg-08.jpg?height=452&width=688&top_left_y=523&top_left_x=685)

## Answer

$\frac{a \sqrt{7}}{12}$.

| Problem 54467 Topics: | Cosine Theorem | ] | Difficulty: 4-  Grades: 8,9 |
| :---: | :---: | :---: | :---: |
|  | [ Equilateral (regular) triangle |  |  |
|  | Median of a trapezoid | ] |  |
|  | Area of a trapezoid | ] |  |