$\underline{\text { Folklore }}$

On the side $B C$ of an equilateral triangle $A B C$, points $K$ and $L$ are marked such that $B K=K L=L C$, and on the side $A C$ a point $M$ is marked such that $A M=1 / 3 A C$. Find the sum of the angles $A K M$ and $A L M$.
Note that triangle $M K C$ is also equilateral, since $C M=\frac{2}{3} C A=2 / 3 C B=C K$ and $\angle M C K=60^{\circ}$ (see the left figure). Therefore, $M K \| A B$, so $\angle A K M=\angle K A B$.

The median $M L$ of triangle $M K C$ is its altitude and, therefore, parallel to the altitude $A D$ of triangle $A B C$; hence, $\angle A L M=\angle L A D$.

$D$ is the midpoint of $K L$, so triangle $K A L$ is isosceles; thus, its altitude $A D$ is also the angle bisector, meaning $\angle L A D=\angle K A D$.

Therefore, $\angle A K M+\angle A L M=\angle K A B+\angle K A D=\angle B A D=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-43.jpg?height=606&width=635&top_left_y=0&top_left_x=277)

## Answer

$30^{\circ}$.

Send a comment