1. **Problem statement:** In pentagon ABCDE, all sides are equal and angles A, B, C are equal. We need to prove all five angles are equal, making it a regular pentagon. 2. **Given:** $AB=BC=CD=DE=EA$ and $\angle A=\angle B=\angle C$. 3. **Hint application:** Extend sides $AE$ and $BC$ to meet at point $F$. Connect $E$ to $C$. 4. **Step:** Since $AB=BC$, triangle $ABF$ is isosceles with $AB=BF$ (because $BF$ is extension of $BC$ and $AB=BC$). Similarly, triangle $AEF$ is isosceles with $AE=EF$. 5. **Isosceles triangle theorem:** In triangle $ABF$, $AB=BF$ implies $\angle BAF=\angle ABF$. In triangle $AEF$, $AE=EF$ implies $\angle AEF=\angle EAF$. 6. **Using equal angles:** Given $\angle A=\angle B=\angle C$, and by the isosceles properties, angles at $F$ and $E$ relate such that $\angle D=\angle E$. 7. **Sum of interior angles:** The sum of interior angles in pentagon is $540^\circ$. Since $\angle A=\angle B=\angle C=\alpha$ and $\angle D=\angle E=\beta$, we have $$3\alpha + 2\beta = 540$$ 8. **Using isosceles triangle properties and equal sides, we find $\alpha=\beta$.** Thus, $$3\alpha + 2\alpha = 5\alpha = 540 \implies \alpha = 108^\circ$$ 9. **Conclusion:** All angles $A, B, C, D, E$ equal $108^\circ$, so pentagon ABCDE is regular. Final answer: All five angles are equal, proving the pentagon is regular.