6. On the side $AB$ of an equilateral triangle $ABC$, a point $K$ is chosen, and on the side $BC$, points $L$ and $M$ are chosen such that $KL = KM$, with point $L$ being closer to $B$ than $M$.

a) Find the angle $MKA$ if it is known that $\angle BKL = 10^{\circ}$.

b) Find $MC$ if $BL = 2$ and $KA = 3$.

Justify your answers for each part.

![](https://cdn.mathpix.com/cropped/2024_05_06_b25cc23802b34308d5ccg-2.jpg?height=465&width=482&top_left_y=887&top_left_x=1501)

![](https://cdn.mathpix.com/cropped/2024_05_06_b25cc23802b34308d5ccg-2.jpg?height=1265&width=1879&top_left_y=1378&top_left_x=91)

| 5 | 6a | 6b | $\Sigma$ |
| :--- | :--- | :--- | :--- |
|  |  |  |  |
# Problem 6.

Answer: a) $130^{\circ} ;$ b) 5 .

First solution. Mark a point $N$ on the segment $MC$ such that $NM = BL$. Since triangle $LKM$ is isosceles, the angles $KLM$ and $KML$ at its base are equal. Therefore, the adjacent angles $KLB$ and $KMN$ are also equal. This implies the equality of triangles $BKL$ and $NKM$ by the second criterion of triangle congruence. Thus, $\angle KNB = \angle KBL = 60^{\circ}$, meaning triangle $BKN$ is also equilateral.

To solve part a), note that from the congruence of the triangles, it follows that $\angle MKN = \angle BKL = 10^{\circ}$. Then $\angle AKM = \angle AKN + \angle NKM = 180^{\circ} - \angle BKN + \angle NKM = 180^{\circ} - 60^{\circ} + 10^{\circ} = 130^{\circ}$. To solve part b)

![](https://cdn.mathpix.com/cropped/2024_05_06_27451585a068edf5f6c5g-2.jpg?height=594&width=636&top_left_y=1839&top_left_x=1292)
it remains to note that $BK = BN$ and $BA = BC$, so $NC = KA$. Then $MC = MN + NC = BL + KA = 2 + 3 = 5$.

Second solution. Draw the altitude $KF$ in triangle $LKM$. Since this triangle is isosceles, this altitude is also the bisector and median. Then, by the sum of the angles in triangle $BKF$, we find that $\angle BKF = 180^{\circ} - \angle KBF - \angle KFB = 30^{\circ}$. Then $\angle LKF = \angle BKF - \angle BKL = 30^{\circ} - 10^{\circ} = 20^{\circ}$. Since $KF$ is the bisector of angle $LKM$, we get that $\angle FKM = \angle FKL = 20^{\circ}$. Then $\angle AKM = 180^{\circ} - \angle BKL - \angle LKF - \angle FKM = 180^{\circ} - 10^{\circ} - 20^{\circ} - 20^{\circ} = 130^{\circ}$.

To solve part b), let the segment $LF$ be denoted by $x$. Then, since $BF$ is the median of triangle $LKM$, $FM = FL = x$. On the other hand, it is known that in a triangle with angles $30^{\circ}, 60^{\circ}, 90^{\circ}$, the side opposite the $30^{\circ}$ angle is half the hypotenuse. Using this fact for triangle $BKF$, we get that $BK = 2 \cdot (x + 2) = 2x + 4$. Using the equality of sides $AB$ and $BC$, we can set up the equation: $3 + 2x + 4 = 2 + 2x + MC$, from which $MC = 5$.