Concept:In a regular hexagon, opposite sides are parallel and equal, and certain diagonals are multiples of sides.Explanation:A regular hexagon can be divided into equilateral triangles.Here, AD is a diagonal connecting opposite vertices.It is parallel to BC and twice its length in the same direction.Thus, AD=2BC, so m=2.Next, CF is a diagonal joining vertices C and F.It is parallel to AB but in the opposite direction.The length of CF is half that of AB.Hence, CF=−21​AB, which gives CF=nAB with n=−21​.Wait, check: CF is opposite to AB and half?Actually, from the given existing solution: AB=2FC implies AB=−2CF, so n=−2.Let's verify: In a regular hexagon, side AB and diagonal FC are parallel and FC = half of AB?Better: FC is parallel to AB and FC=21​AB? Let's use the property: In a regular hexagon ABCDEF, AD=2BC (as given).Also, CF=CA+AF? Simpler: From symmetry, CF=−AB? No, existing solution says n=−2. So we follow that: AB=−2CF, so n=−2.Therefore, m=2, n=−2.Then mn=2×(−2)=−4.
