Problem - 3837
If all sides of a convex pentagon $ABCDE$ are equal in length and $\angle{A}\ge\angle{B}\ge\angle{C}\ge\angle{D}\ge\angle{E}$, show that $ABCDE$ is a regular pentagon.
It is equivalent to showing $\angle{A}=\angle{B}=\angle{C}=\angle{D}=\angle{E}$> If this is not true, then it must have $\angle{A} > \angle{E}$. As shown, both $\triangle{ABE}$ and $\triangle{EAD}$ are isosceles and $AB=AE=DE$. Then the assumption $\angle{A}>\angle{E}$ will mean $BE > AD$. Applying this conclusion on $\triangle{ABD}$ and $\triangle{EBD}$ yields $\angle{BDE} > \angle{DBA}$. Next, because $BC=CD$, we have $\angle{CDB}=\angle{CBD}$. Combining these two means $\angle{D}>\angle{B}$. This is contracting to the given condition of $\angle{B}\ge\angle{D}$. Therefore, it cannot be true that $\angle{A} > \angle{E}$ which implies all the angles are equal.
