In rhombus $A B C D$, points $M$ and $N$ are the midpoints of sides $B C$ and $C D$ respectively. Find the angle $M A N$, if $\angle B A D=60^{\circ}$.
Find the tangent of half the angle $M A N$.

## Solution

Let $Q$ be the point of intersection of the diagonals of the rhombus, $P$ be the point of intersection of the segment $M N$ with the diagonal $A C, A B=a$. Then $M N$ is the midline of the equilateral triangle $B C D$,

$$
\begin{gathered}
M N=\frac{1}{2} B D=\frac{a}{2}, M P=\frac{1}{2} B Q=\frac{a}{4}, \\
A P=A Q+Q P=A Q+\frac{1}{2} A Q=\frac{3}{2} \cdot \frac{a \sqrt{3}}{2}=\frac{3 a \sqrt{3}}{4}, \\
\operatorname{tg} \angle M A P=\frac{M P}{A P}=\frac{a}{4} \cdot \frac{3 a \sqrt{3}}{4}=\frac{\sqrt{3}}{9}, \\
\cos \angle M A N=\cos 2 \angle M A P=\frac{1-\operatorname{tg}^{2} \angle M A P}{1+\operatorname{tg}^{2} \angle M A P}=\frac{1-\left(\frac{\sqrt{3}}{9}\right)^{2}}{1+\left(\frac{\sqrt{3}}{9}\right)^{2}}=\frac{13}{14}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-09.jpg?height=400&width=700&top_left_y=1632&top_left_x=679)

Answer

$\arccos \frac{13}{14}=2 \operatorname{arctg} \frac{\sqrt{3}}{9}$.

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