Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).

Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b323dccc142e96a9907g-32.jpg?height=327&width=495&top_left_y=1316&top_left_x=479)
Answer: 14.

Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=x$. In the right-angled triangle $C K H$ we have $\angle C K H=90^{\circ}-\angle C=\angle A=30^{\circ}$, so $K C=2 \cdot C H=2 \cdot(C M+M H)=$ $2 \cdot(3+x)=6+2 x$. In the right-angled triangle $A B C$ we have $\angle A=30^{\circ}$, so $B C=2 \cdot A C$. Setting up and solving the corresponding equation $31+2 x+3=2 \cdot(4+6+2 x)$, we find $x=7$. Then $L M=2 x=2 \cdot 7=14$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b323dccc142e96a9907g-33.jpg?height=414&width=703&top_left_y=96&top_left_x=379)

Figure 7: to the solution of problem 9.6