Prove: ∠BCD=2∠KPN in Geometric Hexagon

[anemone]

Let $ABCDEF$ be a convex hexagon such that $AB+DE=BC+EF=FA+CD$ and $AB$ // $DE$, $BC // EF$ and $CD$ // $AF$. Let the midpoints of the sides $AF,\,CD,\,BC$ and $EF$ be $M,\,N,\,K$ and $L$ respectively and let $MN$ meets $KL$ at $P$. Show that $\angle BCD=2\angle KPN$.

 

[jvicens]

Thank you for bringing up this interesting problem. I am always excited to explore new mathematical concepts and solve challenging problems.

After carefully examining the given information and drawing a diagram, I have come to the conclusion that the statement is indeed true. Allow me to explain my reasoning below.

First, let us consider the given conditions. We are given that $AB+DE=BC+EF=FA+CD$ and that $AB$ // $DE$, $BC$ // $EF$, and $CD$ // $AF$. This means that the hexagon $ABCDEF$ is equilateral, with all sides equal in length. Additionally, the parallel sides $AB$ and $DE$ form a parallelogram, as do $BC$ and $EF$, and $CD$ and $AF$. This further implies that the diagonals $AD$, $BE$, and $CF$ bisect each other.

Now, let us focus on the midpoints of the sides $AF$, $CD$, $BC$, and $EF$. Since $AF$ and $CD$ are parallel, their midpoints $M$ and $N$ lie on the same line. Similarly, the midpoints $K$ and $L$ of $BC$ and $EF$ also lie on this line. This means that $MN$ and $KL$ are parallel lines.

Next, we can see that the line $MN$ divides the hexagon into two smaller equilateral triangles, $AMN$ and $MNC$. Similarly, $KL$ divides the hexagon into two smaller equilateral triangles, $BKL$ and $KLE$. Since $MN$ and $KL$ intersect at $P$, we can conclude that $P$ is the intersection of the diagonals of the hexagon, and is also the centroid.

Now, let us focus on the angle $\angle KPN$. Since $P$ is the centroid, it divides each median in the ratio $2:1$. This means that $KP:PN=2:1$. Therefore, we can say that $KP$ is twice the length of $PN$. Now, if we draw a line from $K$ to $B$, we can see that it forms a right angle with $PN$, as $KB$ is parallel to $PN$. This means that $KP$ is the hypotenuse of a right triangle $KBP$, with $PN