Elementary Geometry for College Students (5th Edition)

In a regular pentagon all interior angles equal to $108^{\circ}$. $ABE$ and $BCD$ are isosceles triangles, as they have two equal angles (as given in the HINT). If $\angle EAB = 108^{\circ}$, then $\angle ABE= \frac{180^{\circ}-108^{\circ}}{2}= 36^{\circ}$. With the same reasoning $\angle CBD=36^{\circ}$. Now we need to find $\angle EBD$. We know, that $\angle ABE + \angle CBD + \angle EBD = 108^{\circ}$. So $36^{\circ}+36^{\circ}+ \angle EBD =108^{\circ}$ Therefore: $\angle EBD = 108^{\circ}-36^{\circ}-36^{\circ}=36^{\circ}$. $\angle ABE = \angle CBD =\angle EBD = 36^{\circ}$. So the statement is true.