Googlu15 03.10.2016 19:06 Does there exist a hexagon that can be divided into four congruent triangles by a straight cut?
ThE-dArK-lOrD 03.10.2016 19:21 Start with parallelogram $ABCD$ where $AB>BC$, draw segment $BD$ and reflect $\triangle{ABD}$ over $AB$ get $\triangle{ABE}$ and reflect $\triangle{BCD}$ over $CD$ get $\triangle{FCD}$ Consider hexagon $AEBCFD$ and we are done.