The Problems of Ukrainian Authors

Let $I$ be the center of $ABKL$, contextual from the characteristic of the centroid of the triangle and from the condition of the task: $AI=x$, $KI=2x$. Let $BA=x$, then from the characteristic of the bisector $BK=2x$ and $KC=2x$. Then $KALC$ is a trapezium. That's why $KA\parallel LC$ (see Fig. 20). From Thales' theorem $AL=x$. Therefore, in $ABKL$, $KA$ is a bisector by condition and a median because $BA=AL$. Then $ABKL$ is an equilateral triangle, and so if $BK=2x=BL$, therefore it is an equilateral triangle and all its angles are $60^{\circ }$.