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 $AE$ and $BC$ to meet at point $F$. Connect $E$ to $C$. 4. **Step:** Since $AB=BC$, triangle $ABC$ is isosceles with $AB=BC$ and $\angle A=\angle B=\angle C$, so $\triangle ABC$ is isosceles with base $AC$. 5. **Step:** Similarly, $AE=DE$ and $AB=BC=CD=DE=EA$ imply $\triangle AFE$ and $\triangle BCF$ are isosceles. 6. **Step:** By the isosceles triangle theorem, angles opposite equal sides are equal. Since $\angle A=\angle B=\angle C$, and $F$ lies on extensions of $AE$ and $BC$, triangles $AFE$ and $BCF$ share properties that imply $\angle D=\angle E$. 7. **Step:** Using the fact that the sum of interior angles of pentagon is $540^\circ$, and $\angle A=\angle B=\angle C=\alpha$, and $\angle D=\angle E=\beta$, we have: $$3\alpha + 2\beta = 540^\circ$$ 8. **Step:** From the isosceles triangles and equal sides, $\alpha=\beta$. 9. **Step:** Substitute $\beta=\alpha$ into the sum: $$3\alpha + 2\alpha = 5\alpha = 540^\circ \implies \alpha = 108^\circ$$ 10. **Conclusion:** All angles $A, B, C, D, E$ are equal to $108^\circ$, so the pentagon is regular. **Final answer:** All five angles are equal, proving the pentagon is regular.