#### Answer

The statement is true.

#### Work Step by Step

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.