A point inside a quadrilateral

Question

I have deduced the fact that KL = LM by proving that the diagonals of the quadrilateral are equal. BUT, I can't understand how to proceed further and prove the triangle KLM to be equilateral.

/I got the given question from the Facebook page created recently by my maths teacher

Answer 1

Triangle $BOD$ is obtained from $AOC$ with a $120°$ rotation, hence lines $AC$ and $BD$ form angles of $120°$ and $60°$. Segments $KL$ and $LM$, which are parallel to those lines, thus also form an angle of $60°$, which implies that $KLM$ is equilateral.

EDIT. The same by angle chasing.

Let angles $\alpha$, $\beta$ and $\gamma$ in congruent triangles $AOC$ and $BOD$ be defined as in the diagram below. We have then $\alpha+\beta+\gamma+120°=180°$, that is: $\alpha+\beta+\gamma=60°$. On the other hand, as the angles in quadrilateral $COBN$ sum up to $360°$ ($N$ being the intersection of diagonals $AC$ and $BD$) we have: $\angle CNB+\alpha+\beta+\gamma+240°=360°$, whence: $\angle CNB=120°-\alpha-\beta-\gamma=60°$.

But $ML\parallel BD$ and $KL\parallel AC$, hence $\angle KLM=\angle CNB=60°$.

Answer 2

The figure is not unique in a sense that you can raise the point $A$ to get an isosceles trapezium $ABCD$ and still meet all the requirements. Then $\measuredangle KLM=\measuredangle CQB=60$, because they are opposite angles of the parallelogram.