[ Area of a trapezoid $]$ $[$ Ratios of areas $]$

In rhombus $A B C D$ with side $a$, the angle at vertex A is $60^{\circ}$, points $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. Point $K$ lies on side $B C$, segments $A K$ and $E F$ intersect at point $M$. Find $M K$, given that the area of quadrilateral $M K C F$ is $\frac{3}{8}$ of the area of rhombus $A B C D$.
From the equality $S_{\mathrm{BKME}}=\frac{1}{8} S_{\mathrm{ABCD}}$, derive that $B K=\frac{a}{3}$.

## Solution

Since $S_{\mathrm{MKCF}}=\frac{3}{8} S_{\mathrm{ABCD}}$, then

$$
S_{\mathrm{BKME}}=S_{\mathrm{BCFE}}-S_{\mathrm{MKCF}}=\frac{1}{2} S_{\mathrm{ABCD}}-\frac{3}{8} S_{\mathrm{ABCD}}=\frac{1}{8} S_{\mathrm{ABCD}}
$$

If $h$ is the height of the rhombus, then the area of trapezoid $M K B E$ is $\frac{B K+M E}{2} \cdot \frac{h}{2}$. Denoting $B K$ by $x$, we get the equation

$$
\frac{x+\frac{x}{2}}{2} \cdot \frac{h}{2}=\frac{a h}{8}
$$

From this, we find that $x=\frac{a}{3}$. Then

$$
C K=B C-B K=a-\frac{a}{3}=\frac{2 a}{3}
$$

By the cosine rule in triangle $A C K$, we find that

$$
A K^{2}=C A^{2}+C K^{2}-2 C A \cdot C K \cos 30^{\circ}=3 a^{2}+\frac{4 a^{2}}{9}-2 a^{2}=\frac{13 a^{2}}{9}
$$

Therefore,

$$
M K=\frac{1}{2} A K=\frac{a \sqrt{13}}{6}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-44.jpg?height=443&width=809&top_left_y=630&top_left_x=630)

## Answer

$\frac{a \sqrt{13}}{6}$.