$[\quad$ The Law of Cosines

Given an equilateral triangle with side $a$. Find the segment connecting the vertex of the triangle to the point that divides the opposite side in the ratio $2: 1$.

#
Use the Law of Cosines.

## Solution

Let point $M$ lie on side $BC$ of an equilateral triangle $ABC$ with side $a$, such that $BM: CM=2: 1$. Then $CM=\frac{1}{3} a$. By the Law of Cosines in triangle $ACM$, we find that

$$
AM^{2}=AC^{2}+CM^{2}-2 AC \cdot CM \cos 60^{\circ}=a^{2}+\frac{1}{9} a^{2}-\frac{1}{3} a^{2}=\frac{7}{9} a^{2}
$$

Therefore, $AM=\frac{a \sqrt{7}}{3}$.

Answer

$\frac{a \sqrt{7}}{3}$